# Prove whether or not $f(x,y) = \frac{x^2y}{x^2+y^2}$ is differentiable at $(0,0)$.

I have the following function: $$f: \mathbb{R}^2 \to \mathbb{R}$$ with $$f(x,y) = \frac{x^2y}{x^2+y^2}$$ for $$(x,y) \neq (0,0)$$ and $$f(0,0) = 0$$. Now I want to proof that $$f$$ is differentiable in $$(0,0)$$, but I get stuck with an $$\epsilon-\delta$$ proof for the limit $$$$\lim_{(k,h) \to (0,0)} \frac{f(0+k,0+h) - f(0,0) - kf_1(0,0) - hf_2(0,0)}{\sqrt{k^2 + h^2}} = 0$$$$ I think that $$f_1(0,0) = f_2(0,0) = 0$$, because $$\lim_{h \to 0} \frac{f(h,0) - f(0,0)}{h} = \lim_{h \to 0} \frac{\frac{h^2 \cdot 0}{h^2 + 0}-0}{h} = 0$$. And the same for $$y+h$$.

So the actual limit I want to solve is $$$$\lim_{(k,h) \to (0,0)} \frac{\frac{k^2h}{k^2+h^2}}{\sqrt{k^2+h^2}} = \lim_{(k,h) \to (0,0)} \frac{k^2h}{(k^2 + h^2)^{\frac{3}{2}}} = 0$$$$ Perhaps I made a mistake earlier on, but I can't seem to give a right proof of the existence of the limit...

The function $$f$$ is not differentiable at $$(0,0)$$. If it was, then, since $$f_x(0,0)=f_y(0,0)=0$$, then $$f'(0,0)$$ would be the null function. In other words,$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But if $$x=y>0$$, $$\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac1{2\sqrt2}$$.