I know
$$\lim_{(x, y)\to (0,0)} \frac{x^3 + y^4}{x^2 + y^2} = 0$$ so
$$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert \le f(x, y)$$ for some simpler $f(x, y)$ whose limit is also $0$.
How do I find the function $f(x, y)$? In other words, how do I get the upper bound?
Let $x=r\cos{t}, y=r\sin{t}$, then $$\left\lvert \frac{x^3 + y^4}{x^2 + y^2} \right\rvert=|r(\cos^3{t}+r\sin^4{t})|\leq |r||1+r| = \sqrt{x^2+y^2}+x^2+y^2$$
Using the inequalities $x^2+y^2\ge x^2$ and $x^2+y^2\ge y^2$, we can write
$$\begin{align} \left|\frac{x^3+y^4}{x^2+y^2}\right|&\le \frac{|x|^3}{x^2+y^2}+\frac{y^4}{x^2+y^2}\\\\ &\le |x|+y^2 \end{align}$$
