$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, &\mbox{if } (x,y) \neq (0,0)\\ 0, & \mbox{if } (x,y)=(0,0) \end{cases}$ not differentiable in $(0,0)^T$

Question

$f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, &\mbox{if } (x,y) \neq (0,0)\\ 0, & \mbox{if } (x,y)=(0,0) \end{cases}$

I need to show that all partial derivatives in $(0,0)^T$ exist which I already did, but how can I show that $f$ is not differentiable in $(0,0)^T$?

Answer 1

Because it's not even continuous there. See what happens when $x=y$.

Answer 2

Making $y = \lambda x$ we have

$$ \frac{x y}{x^2+y^2} = \frac{\lambda}{1+\lambda^2} $$

then $f(x,y)$ is not even continuous at $(0,0)$

Answer 3

An easy manner to think is using Change of variables like $$ x = R cos \theta $$ and $$ y = R sin \theta $$ Try this!