Show that $f(x,y)=\dfrac{xy^2}{x^2+y^2}$ (with $(x,y)\not=(0,0)$ and $f(0,0)=0$) is continuous but not differentiable at $(0,0)$.
I tried to show continuity with an $\epsilon -\delta$ argument but I don't know how to factorize the expression so that I can have something useful.
For differentiability I think I should show that the partials are not continuous at $(0,0)$. But finding the partials is also painful.
Best Answer
For continuity, a common trick is to express $f(x,y)=g(x,y)h(x,y)$ where $g$ has limit $0$ at $(0,0)$ and $h$ is bounded in a punctured neighborhood of $(0,0)$. This is easy here: $$ g(x,y)=x,\qquad h(x,y)=\frac{y^2}{x^2+y^2} $$ because it's obvious that $0\le h(x,y)\le 1$ for all $(x,y)\ne(0,0)$.
Differentiability doesn't imply continuity of the partial derivatives; in some sense it's the other way round.
This function has partial derivatives at $(0,0)$ and both are zero: $$ \lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h}= \lim_{h\to0}\frac{1}{h}\frac{h\cdot0^2}{h^2+0^2} =0 $$ and $$ \lim_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}= \lim_{h\to0}\frac{1}{h}\frac{0\cdot h^2}{h^2+0^2} =0 $$ So, if the function is differentiable at $(0,0)$, its differential must be zero. Can you go on?