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?
 A: The main reason you're having trouble is the limit doesn't exist. For example, if you approach $(0,0)$ along the line $x = 2y$, you have
$$\begin{equation}\begin{aligned}
\frac{xy(x-y)}{x^4+y^4} & = \frac{(2y)y(2y-y)}{(2y)^4+y^4} \\
& = \frac{2y^3}{17y^4} \\
& = \frac{2}{17y}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
This goes to $\infty$ as $y \to 0^{+}$ and $-\infty$ as $y \to 0^{-}$.
Note a fairly easy way to see this is that the powers of $x$ and $y$ are $4$ in the denominator, but just $3$ in the numerator.
However, for example, if the actual fraction to check in the limit is
$$\frac{xy(x-y)}{x^2+y^2} \tag{2}\label{eq2A}$$
instead, then the limit does go to $0$. You can see this using the squeeze theorem by using the AM-GM inequality, or by just noting $(x - y)^2 = x^2 - 2xy + y^2 \ge 0 \implies x^2 + y^2 \ge 2xy$, plus $(x + y)^2 = x^2 + 2xy + y^2 \ge 0 \implies x^2 + y^2 \ge -2xy$, so when combined it results in $x^2 + y^2 \ge \left|2xy\right| \implies \frac{1}{x^2 + y^2} \le \left|\frac{1}{2xy}\right|$. This gives
$$0 \le \left|\frac{xy(x-y)}{x^2+y^2}\right| \le \left|\frac{xy(x-y)}{2xy}\right|  \le \left|\frac{x-y}{2}\right| \tag{3}\label{eq3A}$$
A: 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 $y<0$).
