show whether $\frac {xy}{x^2+y^2}$ is differentiable in $0$ or not? (multivariable) Q: $f(x,y)=\frac {xy}{x^2+y^2}$ if $(x,y)\not=(0.0)$, and $0$ if $(x,y)=(0,0)$. 
Is $f$ differentiable at $(0,0)$? 
Attempt: $$\lim_{(x,y) \to (0,0)} \frac {f(x,y)-f(0,0)}{||(x,y)-(0,0)||} = \lim_{(x,y) \to (0,0)}  \frac {\frac{xy}{x^2+y^2}}{\frac {\sqrt {x^2+y^2}}{1}}$$
$$=\lim_{(x,y) \to (0,0)} \frac {xy}{(x^2+y^2)^{3/2}}.$$
Let $x=r\cos(\theta)$ and $y=r\sin(\theta)$. Then, 
$$ \frac {xy}{(x^2+y^2)^{3/2}}= \frac {r^2\cos(\theta)\sin(\theta)}{r^{3/2}}=\sqrt r\cos(\theta)\sin(\theta).$$
What should be the next step? If I use polar coordinates, how can I transform $$\lim_{(x,y) \to (0,0)}\rightarrow\lim_{(r,\theta) \to (?,?)}$$??

Hint: Consider the paths $y=mx$, this will show you that your function is not continuous at the origin hence it cannot be differentiable at the origin
