<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] There's a mistake in the first equality: it's false that
$$
|x^2y-xy^2|=x^2|y|-|x|y^2
$$
However, by the triangle inequality,
$$
|x^2y-xy^2|\le x^2|y|+|x|y^2<x^2\sqrt{x^2+y^2}+y^2\sqrt{x^2+y^2}
$$
so
$$
\left|\frac{x^2y-xy^2}{x^2+y^2}\right|
<\frac{(x^2+y^2)\sqrt{x^2+y^2}}{x^2+y^2}=\sqrt{x^2+y^2}
$$
Now the squeeze theorem allows to finish.
</s>