Expert Help with Multivariable Function Questions

Following functions:
$f(x_1,x_2)=\left[\begin{array}{c}{x}_1{x}_2^2+x_1^3{x}_2\\ x_1^2{x}_2+x_1+x_2^3\end{array}\right]$
$g(u)=\left[\begin{array}{c}{e}^u\\ u^2+u\end{array}\right]$
Is it possible to take a derivative of $f(g)$ or $g(f)$. If not - why?

Consider a function $f:Y_n\to <!--\; \to \; -->X_{n^2}$
For any $(a,b)\in <!--\; \in \; -->Y_n$, let $f((a,b))=f(a,b)=an+b.$ Prove that $f$ is a bijection and find its inverse $f^{-<!--\; -\; -->1}$.
How to find an inverse of this multivariable function. I was already able to prove that $f(a,b)=an+b$ is a bijection, but not certain how to find its specific inverse. It would be important to note that $Y_n=X_n\times <!--\; \times \; -->X_n$ where $X_n=\{0,1,2,...,n-<!--\; -\; -->1\}$.

Confusion in integrating multivariable function
Integrate ${\int <!--\; \int \; -->}_C\frac{-<!--\; -\; -->x}{x^2+y^2}dx+\frac{y}{x^2+y^2}dy$
$$C:x=\cos <!--\; \; -->t,y=\sin <!--\; \; -->t,$$ $0\le <!--\; \le \; -->t\le <!--\; \le \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}$
In this case, It's incorrect to integrate it as $\frac{-<!--\; -\; -->1}{2}\ln <!--\; \; -->(x^2+y^2){|}_a^b+\frac{1}{2}\ln <!--\; \; -->(x^2+y^2){|}_c^d$. But,
${\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^1\frac{1}{1-<!--\; -\; -->xy}dxdy={\int <!--\; \int \; -->}_0^1{\int <!--\; \int \; -->}_0^1\frac{-<!--\; -\; -->1}{y}\frac{-<!--\; -\; -->y}{1-<!--\; -\; -->xy}dxdy={\int <!--\; \int \; -->}_0^1\frac{-<!--\; -\; -->1}{y}[\ln <!--\; \; -->(1-<!--\; -\; -->xy{]}_0^1)dy$
When integrands are multivariable functions for both cases, why does only the bottom case work?

How to apply the chain rule to a double partial derivative of a multivariable function?
$$\begin{array}{rl}f(x,y)& =e^{xy}\\ g(x,y)& =f(\sin <!--\; \; -->(x^2+y),x^3+2y+1)\end{array}$$
Let’s compute ${\displaystyle \frac{{\mathrm{\partial <!--\; \partial \; -->}}^{2}g}{\mathrm{\partial <!--\; \partial \; -->}{x}^2}(0,0)}$

In general for a multivariable functions $y(m,n)$, $x(m,n)$,
$$\begin{array}{}\text{(1)}& \frac{dy}{dx}=\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}m}\frac{dm}{dx}+\frac{\mathrm{\partial <!--\; \partial \; -->}y}{\mathrm{\partial <!--\; \partial \; -->}n}\frac{dn}{dx}\end{array}$$
$$\begin{array}{}\text{(2)}& \frac{dx}{dy}=\frac{\mathrm{\partial <!--\; \partial \; -->}x}{\mathrm{\partial <!--\; \partial \; -->}m}\frac{dm}{dy}+\frac{\mathrm{\partial <!--\; \partial \; -->}x}{\mathrm{\partial <!--\; \partial \; -->}n}\frac{dn}{dy}\end{array}$$
so in general clearly $(dy/dx{)}^{-<!--\; -\; -->1}$ need not be equal to $dx/dy$.
Question is what is the precision Mathematical conditions on functions $y(m,n)$ and $x(m,n)$ under which this equality holds.

What is the MacLaurin formula of higher orders for multivariable functions? Example: $f(x,y)=\cos <!--\; \; -->x\cos <!--\; \; -->y$

If the partial derivatives fx and fy of a function f(x, y) are continuous throughout an open region R, then f is differentiable at every point of R.
So in two dimensions, if something is continuous, it might not be differentiable, because it could be pointy. Couldn't that happen in three dimensions too?

$2$ multivariable functions $Q(x,y)$ and $P(x,y)$, I was wondering if finding the point of intersection between these $2$ functions is as easy as making $Q(x,y)=P(x,y)$ as you would do for most single variable functions.
Does the method for finding the intersection of $2$ single variable functions work for multivariable functions?

Find the critical points of $f(x,y)=x^2{e}^y$so far I have found the gradient vector of this function
$▽<!--\; ▽\; -->f=(2x{e}^y,x^2{e}^y)$

Proving that the function given by:
f(x,y)=(sin(x-y))/(x-y) where x≠y and f(x,y) = 1 when x=y is continuous everywhere
Since f is then a combination of continuous single variable functions does that mean the multivariable function is continuous?

Having problems understanding the epsilon-N definition of limits when the function takes multiple variables.
For example, consider the limit $\underset{(x,y)\to <!--\; \to \; -->(\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->})}{lim}x{e}^{-<!--\; -\; -->y}$.
My hunch is that this limit should converge to $0$, as this yields the correct answer and the graph seems to "flatten out" in general when looking far away in the first quadrant.
explain the rigorous definition of limits at infinity? Also, if possible, could you confirm or disprove my guess about $\underset{(x,y)\to <!--\; \to \; -->(\mathrm{\infty <!--\; \infty \; -->},\mathrm{\infty <!--\; \infty \; -->})}{lim}x{e}^{-<!--\; -\; -->y}$?

Verify the continuity of the following multivariable function:
$$f(x,y,z)=\{\begin{array}{ll}\frac{xz-<!--\; -\; -->y^2}{x^2+y^2+z^2}& ,(x,y,z)\ne <!--\; \ne \; -->(0,0,0)\\ \\ 0& ,(x,y,z)=(0,0,0)\end{array}$$
The continuity can be easily rejected by seeing that the function has no limit at origin when we consider it on two paths:
$$(x,0,0),f\to <!--\; \to \; -->0\text{and}(0,y,0),f\to <!--\; \to \; -->-<!--\; -\; -->1$$
As it is clear, this function is homogenous of degree zero also, so:
Can we say all the multivariable functions having the property above has no limit at the origin?
Indeed, we encounter many multivariable functions ${\mathbb{R}}^2\to <!--\; \to \; -->\mathbb{R}$ or ${\mathbb{R}}^3\to <!--\; \to \; -->\mathbb{R}$ and are asked to probe the continuity at the origin.

Generalize Lagrange's mean value theorem for $f:A\to <!--\; \to \; -->{\mathbb{R}}^n$ , differentiable in $n=1$.What I have so far is that if $n=1$, we can write $g_i(t)=f(x_1,..t{x}_i,..,x_n$, where $x_j$ are any real numbers and $x_i$ is $$1$$. We have $g^{\prime }(t)=D_f(t)\cdot <!--\; \cdot \; -->(0,..,1,...0)=\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{x}_i}(t)$, And since $g_i(t):\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ , we can apply the single variable mean value theorem to say that there exists a point $c_{x_i}\in <!--\; \in \; -->[a,b]$ s.t $\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{x}_i}(c_{x_i})=g_i^{\prime }(c_{x_i})=\frac{g_i(a)-<!--\; -\; -->g_i(b)}{a-<!--\; -\; -->b}$, for any partial derivative. However, For any partial derivative we get a different $c_{x_i}$.
Also, if $n>1$, And I`d know how to handle the $n=1$ case, break down $f$ to $n$ different functions, apply the $n=1$ to each of them seperately, but then again, we get $n$ different points $c$ which aren't necessarily the same..
Is this a good way to try and generalize the Lagrange`s Mean Value theorem for multivariable functions?

The function in question is $$f(x,y)=\frac{\sqrt{y-<!--\; -\; -->x^2}}{1-<!--\; -\; -->x^2}$$
The solution:
For the function, we only want real values so $y$ must be larger than or equal to $x^2$. Also, we can't divide by zero, so $x$ cannot be $$1$$.
They say $x$ cannot be one, since this would be division by $$0$$. But what if we took the point $(1,1)$? Then we would have $\frac{0}{0}$, which is indeterminate. So what's to say we couldn't find that the function is defined at $(1,1)$ after all, by taking some kind of limit to the point?

Given a multivariable function $f(x_1,...,x_n)$ and assume that $\frac{\mathrm{\partial <!--\; \partial \; -->}f}{\mathrm{\partial <!--\; \partial \; -->}{x}_i}$ are monotone increasing with respect to $x_i$.

Taylor series of $f(x,y)=x\cdot <!--\; \cdot \; -->e^{\cos <!--\; \; -->xy-<!--\; -\; -->1}$ around the point $(1,0)$, my attempt so far:
Obviously just by checking first and second partial derivatives it's only getting more complicated. I'm not sure how it is possible to determine the general form of $D^{k}f(1,0)((x-<!--\; -\; -->1,y),(x-<!--\; -\; -->1,y)...(x-<!--\; -\; -->1,y))$ (where there are $$k$$ pairs of $(x-<!--\; -\; -->1,y)$. And without it I'm not sure how to determine the infinite series needed in this problem.

Could someone give geometrical interpretation of the meaning of limit and contuinity? What does it mean to say that a limit at a point exists in multivariable function or it does not exist?

How do we know if a multivariable function (or transform) is invertible?

Consider a particle in a three-dimensional potential
$$V(x,y,z)=\frac{A(x^3+2y^3+3z^3+4a^3)}{{(x^2+y^2+z^2+a^2)}^2}$$
This particle has scattering states if $E>E_0$, where
$E_0=$