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...
 A: Hint:
$$\frac{(xy)^3}{x^2+y^6}\le \frac{(xy)^3}{x^2}=xy^3$$
So the limit exists, and equal to ?

Following the response to Marilia bedoya's comment:
If you want to show
$$\lim_{(x,y)\to(0,0)}\frac{xy^3}{x^2+y^6}$$ doesn't exist,
it suffices to select two paths such that $(x,y)\to (0,0)$ but gives different limits.
Say 1)$x=0,y\to 0$, the limit is $0$. 2)$x=y^3, y\to 0$, the limit is $\frac{1}{2}$.
A: 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:
$$\lim_{r\to 0}\dfrac{r^3\cos^3\theta \cdot r^3\sin^3\theta}{r^2\cos^2\theta+r^6\sin^6\theta}=\lim_{r\to 0}\dfrac{r^3(\cos^3\theta\sin^3\theta)}{r^2(\cos^2\theta+r^4\sin\theta)}=\lim_{r\to 0}\dfrac{r(\cos^3\theta\sin^3\theta)}{(\cos^2\theta+r^4\sin\theta)}=0$$
A: (This answers the question as posted in the headline and the body of the text as of 11/16/14. It seems that the OP had something different in mind.)
Put
$$f(x,y):={(xy)^3\over x^2+y^6}\qquad\bigl(r:=\sqrt{x^2+y^2}\ne0\bigr)\ .$$
Claim: $\qquad\qquad \lim_{(x,y)\to(0,0)}f(x,y)=0\ .$
Proof: We may assume $x\geq0$, $y\geq0$. When $0\leq x\leq y^3$
then $y>0$ and
$$0\leq f(x,y)\leq {(y^3\>y)^3\over y^6}=y^6\leq r^6\ .$$
When $0\leq y^3\leq x$ then $x>0$ and
$$f(x,y)\leq {x^3\>x\over x^2}=x^2\leq r^2\ .$$
It follows that $|f(x,y)|\leq r^2$ as soon as $0<r\leq1$, which proves the claim.
A: If the correct wording is 
$$\lim_{(x,y)\rightarrow (0,0)} \frac{x(y^3)}{x^2+y^6}$$
it is a good idea to change into :
$a=x$, 
$b=y^3$
then is there a "limit" to :
$$\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}$$
$$\frac{ab}{a^2+b^2}=\frac{1}{\frac{a}{b}+\frac{b}{a}}$$
if $a$ and $b$ both tend to $0$ then $\frac{a}{b}$ and $\frac{b}{a}$ are undetermined such as $\frac{0}{0}$. So a limit doesn't exists. 
$\frac{ab}{a^2+b^2}$ is undeterminated in $a$ and $b$ tending to $0$.
You obtain as many different values as you want, depending the manner of making $a$ and $b$ tend to $0$. For examples :
If $a=b$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{1}{2}$
If $a=2b$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{2}{5}$
If $a=kb$ tend to 0, then $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{k}{1+k^2}$
