I am trying to understand the concept of differential of multi-variable functions with which I have the following function:
$$f(x)=\left\{\begin{array}{cc} \frac{2x^2-y^2}{x^2+2y^2} & (x,y) \ne (0,0) \\ 5 & (x,y) = (0,0) \end{array}\right. $$
Whay I have done is, to find the partial derivatives $\frac{\partial f}{\partial{x}}$ and $\frac{\partial f}{\partial{y}}$ of the function for $(x,y) \ne (0.0)$ then, find the alternated limits $\lim_{x\to0}{(\lim_{y \to 0}{(\frac{2x^2-y^2}{x^2+2y^2})})}$ and $\lim_{y\to0}{(\lim_{x \to 0}{(\frac{2x^2-y^2}{x^2+2y^2})})}$ and once they exist, compare them and if they are equal then the function is differentiable at the point $(0,0)$?
Would that be enough to prove that the function is differentiable at the point? What am I missing?
The function is not continuous at the origin and thus is not differentiable there. The way I first saw this was by enforcing the change of variables $$ (x,y)\mapsto \left(r\cos \theta,\frac{r}{\sqrt{2}}\sin\theta\right) $$ Then your limit becomes $$ \lim_{r\to 0}\frac{2r^2\cos\theta-\frac{r^2}{2}\sin^2\theta}{r^2}\\ =2\cos\theta-\frac{1}{2}\sin^2\theta $$ as we shrink the ellipses major and minor radii to $0$, which is not independent of $\theta$, as is required for continuity.
However, if this makes you uncomfortable, it might be easier to take $y=x$ and find the limit at $0$ is $1/3$ while instead, if $y=x^2$, the limit is $2$.