Evaluate $\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2}$ 
Evaluate the following limit: $$\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2}$$


My try:
Using polar substitution - putting $x =r\cos(\theta)$, $y = r\sin(\theta)$,
$$\implies\lim_{r \to 0} \frac{-r^2\cos(\theta) \sin(\theta)}{r^2(\sin^2\theta+\cos^2\theta)} =\lim_{r \to 0} -\cos(\theta) \sin(\theta) = -\cos(\theta) \sin(\theta)$$
Does this make any sense?
Also, approaching from two different sides,
$$\begin{align}\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2} & = \lim_{x\to 0}\frac{-x(0)}{x^2 + (0)^2} = 0\\\lim_{(x,y) \to (0,0)}\frac{-xy}{x^2 + y^2} & = \lim_{y\to 0}\frac{-(0)y}{(0)^2 + y^2} = 0 \end{align}$$
Both are equal... but the limit doesn't exist which I verified from Symbolab. What should I do to evaluate this?
 A: Consider $(x, y) = (\frac{1}{n}, \frac{1}{n})$ as $n \to \infty$. In that case, your limit evaluates to $-\frac{1}{2}$, which is a contradiction with the other two limits you found so the limit doesn't exist.
A: Let $y=ax$. Then
$$
\lim_{x\to0}f(x,ax)=\frac{-a}{1+a^2}
$$
and certainly depends on $a$.
A: Using polar coordinates you find that $$\lim_{(x,y)\to (0,0)}\frac{-xy}{x^{2}+y^{2}}=\lim_{r\to 0;r>0}\frac{-r^{2}\cos \theta\sin \theta}{r^{2}}=-\cos\theta\sin\theta, \quad (*)$$
So the limit doesn't exists, it depends of values for $\theta$ , just take $\theta=\pi$ and we have $0$ as limit but $\theta=\pi/6$ we have $-\sqrt{3}/4$ as limit, contradiction. I think it's perfectly valid argument. If you need use path, just see in the path $r(t)=(t,mt)$ with $m\in \mathbb{R}^{*}$.
EDIT:
As pointed out Hans Lundmark in the comment, the line in $(*)$ has problems about rigor mathematics. We will try to fix with change of variables to polar coordinates we have for $x^{2}+y^{2}\not=0$, that,
$$\frac{-xy}{x^{2}+y^{2}}=-\frac{r^{2}\cos\theta\sin\theta}{r^{2}}=-\cos\theta\sin\theta, \quad r>0$$
and since for different values of $\theta$ we get differents values so the $\displaystyle \lim_{(x,y)\to (0,0)}\frac{-xy}{x^{2}+y^{2}}$ doesn't exists.
