<s>[INST] 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...
 [/INST] 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 $\lim_{(a,b)\rightarrow (0,0)} \frac{ab}{a^2+b^2}=\frac{k}{1+k^2}$
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