Proving that $\frac{xy^2}{x^2+y^4}$ is bounded I am trying to prove that there exists $M>0$ such that, for all $(x,y) \in \mathbb{R}^2$,
$|\frac{xy^2}{x^2+y^4}|\leq M$. Is the following proof correct?
I claim that there exists such an $M$, and that $M=1$.
Suppose, first, that $|x^2|\geq |xy^2|$. Then $|\frac{xy^2}{x^2+y^4}|\leq |\frac{x^2}{x^2+y^4}|\leq |\frac{x^2}{x^2}|=1$.
Suppose, second, that $|x^2|<|xy^2|$. Then $|x|<|y^2|$, so $|xy^2|<|y^4|.$ Then $|\frac{xy^2}{x^2+y^4}|\leq |\frac{y^4}{x^2+y^4}|\leq |\frac{y^4}{y^4}|=1$ (note that if $|x|=0$, then $f(x)=0$—thus our assumption that $|x|>0$ is not without loss of generality).
Thus, for any $(x,y) \in \mathbb{R}^2$, $|f(x,y)|\leq M$

Your proof is a good attempt, but it is not complete. The "mistake" in the proof is that the inequality in red below is not justified:

Suppose, second, that $x^2<xy^2$. Then $|\frac{xy^2}{x^2+y^4}|\color{red}{\leq} |\frac{y^4}{x^2+y^4}|\leq |\frac{y^4}{y^4}|=1$.

You do not explain why this is true. Your assumption is that $x^2<xy^2$, but then you somehow conclude that $|xy^2| \leq |y^4|$. Why is this so?

To be more rigorous, you could argue like so:

If $x^2 < xy^2$, then $x>\frac{x^2}{y^2}\geq 0$. Therefore, we can divide the inequality by $x$ and get $x<y^2$, meaing that $0<x<y^2$.
This means that $|xy^2| = xy^2 < (y^2)y^2 = y^4=|y^4|$.

