Evaluating $\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$ $$\lim_{(x,y)\rightarrow (0,0)} \frac{(xy)^3}{x^2+y^6}$$
I don't really know how to do, but I was trying to do like that:
$a=x$, 
$b=y^2$
then I was trying to do this
$$\lim_{(x,y)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
then I don't know no more how to do...

Conversion into polar coordinates can be of help as well. Letting $x=r\cos\theta$ and $y=r\sin\theta$, we can write the limit as follows:
$$\lim_{r\to 0}\dfrac{r^3\cos^3\theta \cdot r^3\sin^3\theta}{r^2\cos^2\theta+r^6\sin^6\theta}=\lim_{r\to 0}\dfrac{r^3(\cos^3\theta\sin^3\theta)}{r^2(\cos^2\theta+r^4\sin\theta)}=\lim_{r\to 0}\dfrac{r(\cos^3\theta\sin^3\theta)}{(\cos^2\theta+r^4\sin\theta)}=0$$
