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