Proving that $\lim_{(x;y)\rightarrow (0;0)} \frac{xy(x-y)}{x^4+y^4} = 0$ using the squeeze theorem My attempt:
$$0 \le \frac{|xy(x-y)|}{|x^4+y^4|} = \frac{|x^2y-xy^2|}{|x^4+y^4|} \le \frac{|x^2y|+|xy^2|}{x^4+y^4} \le \frac{|x|^2|y|+|x||y|^2}{x^4+y^4} = \frac{|x|^2|y|+|x||y|^2}{(x^2+y^2)^2-(\sqrt{2}xy)^2} = ?$$
What do I do next?

This limit is not $0$ (I bet it doesnt even exist).
If we move towards $0$ on the line $2x=y$ we have
$$\frac{xy(x-y)}{x^4+y^4}=\frac{2y^3}{17y^4}=\frac{2}{17y}$$
and as $y\to0$ this is $\pm\infty$ (depending on whether $y>0$ or $y<0$).
