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 $x^2-y^2$

The first inequality you're using has a flaw, as it is : $ |xy|\leqslant x^2+y^2$ and after that you can continue on. 
In my answer,I assume that you have made a typo and in the denominator you have $y^2$, since I see you using it in the rest of your attempt.
Showing you some approaches here :
Re-writing the Limit using polar coordinates $x = r\sin\theta,y=r\cos\theta$, we get :
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0 \Rightarrow \lim_{r\to0} \frac{(r\cos\theta)^2(r\sin\theta)-(r\sin\theta)^2(r\cos\theta)}{r^2} \Rightarrow \lim_{r\to0} \frac{r^3\cos^2\theta\sin\theta-r^3\sin^2\theta\cos\theta}{r^2} \Rightarrow \lim_{r\to0} r(\cos^2(\theta)\sin(\theta)-\sin^2(\theta)\cos(\theta)) =0$$
In this case, the value of $\theta$ is insignificant.
In another approach, you have $\forall (x,y)\neq (0,0)$:  
$$\left|\frac{xy}{x^2+y^2}\right|\leq 1 $$
So :
$$\left|\frac{xy(x-y)}{x^2+y^2}\right|\leq |x-y|\leq |x|+|y|\leq 2\sqrt{x^2+y^2}.$$
Now, for any $ε>0$ defining it as $δ=\frac{ε}{2}$, we have :
$$\sqrt{x^2+y^2}<\delta \Rightarrow \left|\frac{xy(x-y)}{x^2+y^2}\right|<\epsilon.$$
