Prove $f(x,y)$ is not differentiable in ($0,0)$ Prove $f(x,y)$ is not differentiable in ($0,0)$
$$f(x,y)= \begin{cases}\dfrac{|x|y}{\sqrt{x^2+y^2}}& \text{if } (x,y)\not =0\\ \\ 0&\text{if } (x,y)=0
\end{cases}
$$
I try prove this by existence of the limit.
Let $y=x$ and $x\not = 0$, then
$$f(x,y)=\frac{|x|x}{\sqrt{x^2+x^2}}=\frac{|x|x}{x\sqrt{2}}=\frac{|x|}{\sqrt{2}}$$
And $\lim_{x\rightarrow 0}f(x,y)=0.$
Moreover, let $x=my$ then
$$f(x,y)=\frac{|my|y}{\sqrt{m^2y^2+y^2}}=\frac{m|y|y}{\sqrt{y^2(m^2+1)}}=\frac{m|y|}{\sqrt{(m^2+1)}}$$
This implies $\lim_{y\rightarrow 0}f(x,y)=0.$
I try other trayectories but don't work. Can someone help me with this?

Check from the definition that the partial derivatives of $f$ at $(0,0)$ are $0.$ Thus if $Df(0,0)$ exists, it equals the $0$ linear transformation. It would then follow that all directional derivatives at $(0,0)$ are $0.$ But 
$$\frac{f(x,x) -f(0,0)}{x} = \frac{|x|x/(\sqrt 2 |x|)}{x} = \frac{1}{\sqrt 2},$$
which does not $\to 0,$ contradiction. Hence $Df(0,0)$ does not xist.
