Limits with variable substitution and Trig \lim_{x\to0}\frac{\sin^2(3x)}{x^2\cos x}=\lim_{u\to0}\frac{\sin^2(u)}{(\frac u3)^2}=9(\lim_{u\to0}\frac{\sin u}u)^2

Proof: There is no function $f\in C^2({\mathbb{R}}^{\mathbb{3}})$ with gradient $\mathrm{\nabla }f(x,y,z)=(yz,xz,x{y}^2)$
How can one show that there is no function, which is a continuously partially derivable function $f\in C^2({\mathbb{R}}^{\mathbb{3}})$ with this gradient $\mathrm{\nabla }f(x,y,z)=(yz,xz,x{y}^2)$.
I thought about using the Hessian matrix since one has to calculate all second partial derivatives of f there.
Since only the gradient is given, can I calculate the antiderivatives first:
$yz=xyz$
$xz=xyz$
$x{y}^2=x{y}^{2}z$
Now I want to calculate the antiderivatives of the antiderivatives:
$xyz=\frac{yz{x}^2}{2}$
$xyz=\frac{yz{x}^2}{2}$
$x{y}^{2}z={\displaystyle \frac{y^{2}z{x}^2}{2}}$
I didn't calculate the antiderivatives of the partial derivatives and I don't even know if that way is correct...

What is the equation of the normal line of ${\displaystyle f\left(x\right)=-x^4+4x^3-x^2+5x-6}$ at x=2?

I'm trying to show
$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{(\sqrt{n}+n-1)!}{(\sqrt{n})!(n-1)!}}{2^{\sqrt{n}}}=\mathrm{\infty }$

How can I calculate the following difficult limit: ${\displaystyle \underset{x\to \mathrm{\infty }}{lim}(x+\sin x)\sin \frac{1}{x}}$ ?

Determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided limits are ${\displaystyle \mathrm{\infty }}$ or ${\displaystyle -\mathrm{\infty }}$, and describe the corresponding vertical asymptote.
${\displaystyle \underset{x\to -3^{\pm }}{lim}\frac{x^2}{x^2-9}}$

If mn$\frac{m}{n}$ is an approximation to $\sqrt{2}$ then prove that $\frac{m}{2n}+\frac{n}{m}$ is a better approximation to $\sqrt{2}.$.(where mn is a rational number)