Find $\lim_{(x,y)\to(0,0)}\frac{-x(\ln(1+y^2)+y^2)}{(x^2+y^2)^\frac{3}{2}}$ 
Find $\lim_{(x,y)\to(0,0)}\frac{-x(\ln(1+y^2)+y^2)}{(x^2+y^2)^\frac{3}{2}}$

I have tried approaching different paths to show that maybe it does not exist but I am really stuck.. appreciate any help!
P.S I have plugged this into wolfram alpha and it said that limit does not exist
 A: Let $y=mx$
Then you get $$\frac{x(\ln(1+m^{2}x^{2})+m^{2}x^{2})}{x^{2}(1+m^{2})^{\frac{3}{2}}}=\frac{x\ln(1+m^{2}x^{2})+m^{2}x^{3}}{(1+m^{2})^{\frac{3}{2}}x^{3}}$$
Now use the fact that $\displaystyle\frac{x\ln(1+m^{2}x^{2})}{x^{3}}=\frac{\ln(1+m^{2}x^{2})}{x^{2}}\xrightarrow{x\to 0} m^{2}$ to see that
for $y=mx$ the limit equals $\displaystyle\frac{-2m^{2}}{(1+m^{2})^{\frac{3}{2}}}$ which is different for different $m$.
A: In this case, polar coordinates reveal clearly the non-existence of the limit.
As usual $x=r\cos t,\; y=r\sin t$:
\begin{eqnarray*} -\frac{x(\ln (1+y^2) + y^2)}{(x^2+y^2)^{\frac 32}}
& = & -\frac{\cos t}{r^2}\left(\ln (1+r^2\sin^2 t) + r^2\sin^2 t\right) \\
& \stackrel{\sin t \neq 0}{=} & -\cos t \sin^2 t \left(\underbrace{\frac{\ln (1+r^2\sin^2 t)}{r^2\sin^2 t}}_{\stackrel{r\to 0}{\longrightarrow}1} + 1\right)
\end{eqnarray*}
So, obviously the behavior of the expression for $r \to 0$ depends only on $t$ and different values like $t= \frac{\pi}2$ or $t= \frac{\pi}4$ produce different results.
