show whether $\frac {xy}{x^2+y^2}$ is differentiable in $0$ or not? (multivariable)

Question

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 $$\lim_{(x,y) \to (0,0)}\rightarrow\lim_{(r,\theta) \to (?,?)}$$??

Answer 1

$$ f\left(\frac{1}{n},\frac{1}{n}\right)=\frac{1}{n^2}\frac{n^2}{2}=\frac{1}{2} $$ And $\displaystyle \left(\frac{1}{n},\frac{1}{n}\right) \underset{n \rightarrow +\infty}{\rightarrow}(0,0)$, so it is not continuous at $(0,0)$. It cannot be differentiable.

Answer 2

Hint:

AM-GM gives

$$\left|\frac{xy}{x^2+y^2}\right| = \frac{|x||y|}{x^2+y^2} \le \frac{\frac{x^2+y^2}2}{x^2+y^2} = \frac12$$

with equality when $x=y$.

Therefore $\lim_{(x,y)\to (0,0)} f(x,y) \ne 0$ so $f$ is not continuous at $(0,0)$.

Answer 3

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