I need to prove that
$$ \lim_{(x,y)\to(0,0)}(x^2-y^2)\sin(\frac{1}{x^2+y^2})=0$$
But I'm not sure about my demonstration. I see that $\left|\sin(\frac{1}{x^2+y^2})\right| \le 1$ So I write that
$$\lim_{(x,y)\to(0,0)}(x^2-y^2)\sin(\frac{1}{x^2+y^2})\le\lim_{(x,y)\to(0,0)}\left|(x^2-y^2)\sin(\frac{1}{x^2+y^2})\right| \le |(x^2-y^2)|\le|x^2|+|y^2|\le-|x^2|\underbrace{\to0}_{x\to 0}$$
By the Squeeze Theorem
$$\lim_{(x,y)\to(0,0)}(x^2-y^2)\sin(\frac{1}{x^2+y^2})\le\lim_{x\to0}-|x^2|=0$$
All that you have proved was that if your limit exists, then it cannot be greater than $0$. You can prove that it is $0$ using the fact that$$\left\lvert(x^2-y^2)\sin\left(\frac1{x^2+y^2}\right)\right\rvert\leqslant x^2+y^2.$$