Proving $ \frac{x^3y^2}{x^4+y^4}$ is continuous. [duplicate]

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 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.

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.