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?

If it were differentiable at $(0,0)$, the corresponding Jocobian is the partial derivatives at $(0,0)$, which you can compute the Jacobian is actually the zero map as well. 
So you may try to get a contradiction with the existence of 
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\dfrac{1}{\sqrt{x^{2}+y^{2}}}\dfrac{|x|y}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\rightarrow(0,0)}\dfrac{|x|y}{x^{2}+y^{2}}.
\end{align*}
For $y=mx$, $x>0$, then
\begin{align*}
\lim_{x\rightarrow 0^{+}}\dfrac{mx^{2}}{x^{2}+m^{2}x^{2}}=\lim_{x\rightarrow 0^{+}}\dfrac{m}{1+m^{2}}=\dfrac{m}{1+m^{2}},
\end{align*}
which varies along with $m$.
