Showing that a function from $\mathbb{R}^2 \to\mathbb{R}$ is bounded 
Show that 
  $$ f(x,y)= \begin{cases} \dfrac{xy^2}{x^2+y^4} &  (x,y) ≠ (0,0) \\ 0 & (x,y) = (0,0) \end{cases}$$
  is bounded.

I thought about splitting it up into different cases like $x<y$ but it turned out to be too many and I could not cover all of them.
As a hint I got the idea to use the arithmetic geometric inequality.
I hope someone can help me. 

The insight here is that the fraction is of the form $\frac{ab}{a^2+b^2}$, with $a=x$ and $b=y^2$.
\begin{align*}
    (x-y^2)^2 &\geq 0 \\
\implies x^2 - 2xy^2 + y^4 &\geq 0 \\
\implies x^2 + y^4 &\geq 2xy^2 \\
\implies\frac{xy^2}{x^2 + y^4} &\leq \frac{1}{2}
\end{align*}
You can also apply the arithmetic mean-geometric mean inequality with $a = x^2$ and $b=y^4$.  Then
\begin{align*}
    \sqrt{ab} &\leq \frac{1}{2}(a + b) \\
    \implies xy^2 &\leq \frac{1}{2}(x^2+y^4) \\
    \implies \frac{xy^2}{x^2+y^4} &\leq \frac{1}{2}
\end{align*}
