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.
 A: To expand on my comments, let $\epsilon > 0 $ be given, then our limit, in polar coordinates is of the form
$$f(r,\theta)={r^5\cos^3\theta\sin^2\theta\over r^4(\cos^4\theta+\sin^4\theta)}$$
this satisfies
$$|f(r,\theta)-0|\le |r|\left|{\cos^3\theta\sin^2\theta\over \cos^4\theta+\sin^4\theta}\right|\le {r\over m_\theta}={\delta\over m_\theta}$$
with $m_\theta$ the minimum value of $\cos^4\theta+\sin^4\theta$ for $0\le\theta<2\pi$. What is this minimum?
Taking derivatives, we see that it is possible when $-4\cos^3\theta\sin\theta+4\sin^3\theta\cos\theta=0\quad (*)$
three cases
(i)  $\theta=0,\pi$ then the function's value is $1$.
(ii)  $\theta=\pi/2, 3\pi/2$ again, value $1$.
(iii)  $\cos\theta\sin\theta\ne 0$ then we have $\sin^2\theta=\cos^2\theta$ by manipulating $(*)$. This only happens when $\theta$ is on the line $y=\pm x$, whence both $\sin^4\theta$ and $\cos^4\theta$ are positive, so the minimum is a positive number, in fact it's easily seen to be $m_\theta={1\over 2}$
Hence we choose $\delta={\epsilon\over 2}$ and we are guaranteed convergence to $0$, as claimed.
A: 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.
