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

