Does the limit $\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}$ exists? I was trying to find if the the following limit exists:
$$
\lim_{(x,y)\to (0,0)} \frac{x^3y^2}{x^6+y^4}
$$
From this topic I tried to switch to polar coordinates:
$$
\frac{x^3y^2}{x^6+y^4}=\frac{r^3\cos^3\theta\cdot r^2\sin^2\theta}{r^6\cos^6\theta+r^4\sin^4\theta}=\frac{r\cos^3\theta\sin^2\theta}{r^2\cos^6\theta+\sin^4\theta}\to 0
$$
But apparently this limit does not exists (proved it in another way). Why this method worked for the limit in that topic and here it didn't work?

The conversion to polar coordinates has not failed.  Note what happens when $r=\tan^2(\theta)\sec(\theta)$, $|\theta| > \pi/2$.  Then,
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=\frac{\sin^4(\theta)}{2\sin^4(\theta)}=\frac12$$
And of course, if $\theta =0$ or $\theta =\pi/2$, then
$$\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}=0$$

EXPLANATION:  So what is happening here?
Note that $x^6+y^4\ge 2\sqrt{x^6y^4}=2|x|^3y^2$, with equality for $y^2=|x|^3$. Therefore, we have
$$\left|\frac{x^3y^2}{x^6+y^4}\right|\le \frac12$$
with equality when $y^2=|x|^3$.
If we transform to polar coordinates, this analysis is identical to
$$\left|\frac{r\cos^3(\theta)\sin^2(\theta)}{r^2\cos^6(\theta)+\sin^4(\theta)}\right|\le \left|\frac{r\cos^3(\theta)\sin^2(\theta)}{2\sqrt{r^2\cos^6(\theta)\sin^4(\theta)}}\right|=\frac12$$
with equality when $r=\frac{\sin^2(\theta)}{|\cos^3(\theta)|}=\tan^2(\theta)|\sec(\theta)|$.
