How do you prove Euler's Theorem du=(du/dx)dx+(du/dy)dy if u=f(x,y). I also heard that Ramanujan developed another method can somebody know that?

How do you prove Euler's Theorem
$du=\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\right)dx+\left(\frac{\mathrm{\partial }u}{\mathrm{\partial }y}\right)dy$
if $u=f\left(x,y\right)$.
I also heard that Ramanujan developed another method can somebody know that?
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phinny5608tt
Let $p=\left(x,y\right)$ and $f:\mathrm{\Omega }\subseteq {\mathbb{R}}^{2}\to \mathbb{R}$ be a function defined on the open set $\mathrm{\Omega }\ni p$.
We say that $f$ is differentiable at $p$ if for every $h:=\left(dx,dy\right)$ s.t. $p+h\in \mathrm{\Omega }$ there exists a vector $a\in {\mathbb{R}}^{2}$ s.t. $df\left(p\right):{\mathbb{R}}^{2}\to \mathbb{R}$ s.t.
$f\left(p+h\right)-f\left(p\right)=⟨a,h⟩+O\left(‖h{‖}^{2}\right).$
The linear map $a\to ⟨a,h⟩$ from ${\mathbb{R}}^{2}$ to $\mathbb{R}$ is called differential of $f$ at $p$ and it is denoted by $df\left(p\right)$. The vector $h$ can be consideblack as a small increment around $p$.
If $f$ is differentiable it follows that (this is shown on every textbook of Analysis)
$df\left(p\right)\left(a\right):=⟨a,h⟩=⟨\mathrm{\nabla }f\left(p\right),h⟩,$
where $\mathrm{\nabla }f\left(p\right)$ is the gradient of $f$ at $p$. In summary, recalling that $p=\left(x,y\right)$ and $h=\left(dx,dy\right)$ we arrive at
$df\left(p\right)=⟨\mathrm{\nabla }f\left(p\right),h⟩=\frac{\mathrm{\partial }f}{\mathrm{\partial }x}dx+\frac{\mathrm{\partial }f}{\mathrm{\partial }y}dy,$
with the partial derivatives computed at $p$.