Continuity of $\frac{x^3y^2}{x^4+y^4}$ at $(0,0)$? [duplicate]

Suppose a function $f$ is defined as follows:

$$f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}$$

Is this function continuous at $(0,0)$? How is this shown? I've tried considering limits for different $y=g(x)$ functions and I am unable to find a counterexample. But I do not see how to prove continuity in general.

Since $x^4-2x^2y^2+y^4= (x^2-y^2)^2 \ge 0$, we have $2x^2y^2 \le x^4+y^4$.

Therefore, $\dfrac{x^2y^2}{x^4+y^4} \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$.

Also, $\dfrac{x^2y^2}{x^4+y^4} \ge 0$ for all $(x,y) \neq (0,0)$.

From the above inequalities, we have that $\left|\dfrac{x^2y^2}{x^4+y^4}\right| \le \dfrac{1}{2}$ for all $(x,y) \neq (0,0)$

Now, multiply both sides by $|x|$ to get $\left|\dfrac{x^3y^2}{x^4+y^4}\right| \le \dfrac{1}{2}|x|$.

Can you finish the problem from here?

The Squeeze Theorem will be useful.

A fairly efficient way to approach this problem is to transform to polar coordinates and write

$$\lim_{(x,y)\to (0,0)}\frac{x^3y^2}{x^4+y^4}=\lim_{r\to \infty}\left(r\,\,\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right)$$

Noting that we can write

$$\begin{align} \left|\frac{\cos^3(\phi)\sin^2(\phi)}{\cos^4(\phi)+\sin^4(\phi)}\right|&=\left|\frac{\cos^3(\phi)\sin^2(\phi)}{2(\sin^2(\phi)-\frac12)^2+\frac12}\right|\\\\ &\le 2 \end{align}$$

then the limit of interest is $0$. Therefore, the function $f(x,y)$ is continuous at the origin.