Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. The problem asks to show that $$f(x,y) = \left\{ \begin{align} \frac{x^3y^2}{x^4+y^4}, & (x,y) \neq (0,0), \\ 0, & (x,y) = (0,0), \end{align} \right.$$ is continuous at the origin, however it has resisted my bravest efforts. I have attempted using $x^4+y^4 \geq y^4$ and therefore $$\left\vert \frac{x^3y^2}{x^4+y^4} \right\vert \leq \left\vert \frac{x^3}{y^2} \right\vert$$ and similar strategies but they have failed. Trying to disprove it and see if it's discontinuous has only strengthened the belief that it's continuous.

To apply the squeeze theorem, notice that if $x\neq 0$ and $y\neq 0$ then
$$\begin{align*}
\left|\frac{x^3y^2}{x^4+y^4}\right|&=\left|x\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
&=\left|x\frac{x^{-2}}{x^{-2}}\frac{x^2}{\sqrt{x^4+y^4}}\frac{y^{-2}}{y^{-2}}\frac{y^2}{\sqrt{x^4+y^4}}\right|\\
& =\left|x\frac{1}{\sqrt{1+y^4/x^4}}\frac{1}{\sqrt{x^4/y^4+1}}\right|\\
&=|x|\left|\frac{1}{\sqrt{1+y^4/x^4}}\right|\left|\frac{1}{\sqrt{x^4/y^4+1}}\right|\\
&\leq |x|\cdot 1\cdot 1.
\end{align*}$$
So, for all $(x,y)\in\mathbb{R}^2$ we have $0\leq |f(x,y)|\leq|x|$. It follows that
$$\lim_{(x,y)\to(0,0)}f(x,y)=0=f(0,0)$$
and thus $f$ is continuous at origin.
