<s>[INST] prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that 
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y^2}<\frac{x^2 \sqrt{x^2+y^2} -y^2\sqrt{x^2+y^2}}{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2-y^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2-y^2) }{x^2+y^2}=\frac{\sqrt{x^2+y^2}(x^2) }{x^2+y^2}<\frac{\sqrt{x^2+y^2}(x^2+y^2) }{x^2+y^2}=\sqrt{x^2+y^2}<\delta=\epsilon$$
I use that $x^2<x^2+y^2$ twice, can I do it? or some identity to use when you have $x^2-y^2$
 [/INST] If the question is correct:
$f(x,y) = \frac {xy^2 - x^2y}{x^2 + y}$
Suppose $y = -x^2 + \zeta$
where $\zeta$ is something very small.
$f(x,y) \approx  \frac{x^2}{\zeta}$
There exists an $(x,y)$ in a neighborhood of $(0,0)$ such that $f(x,y)$ is large.
The limit does not exist.
If it is 
$f(x,y) = $$\frac {xy^2 - x^2y}{x^2 + y^2}\\
\frac {xy(x-y)}{x^2 + y^2}$
$2xy < x^2+y^2 $ by the AM-GM inequality
$f(x,y) < \frac 12 (x-y)$
and $|f(x,y)| < \frac 12 (|x| + |y|)$
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