Show that $f(x,y)=\frac{xy^2}{x^2+y^4}$ is bounded 
Let $f\colon\mathbb R^2\to\mathbb R$ be a function given by:
  $$
f(x,y)=\begin{cases}\frac{xy^2}{x^2+y^4}&\text{if }(x,y)\neq(0,0),\\
0&\text{if }(x,y)=(0,0).
\end{cases}
$$

I need to show that $f$ is bounded on $\mathbb R^2$, so I need to show that there exists $M>0:\vert f(x,y)\vert\leq M$ for all $(x,y)\in\mathbb R^2$.
We can rewrite $\begin{align}f(x,y)=\frac{x/y^2}{1+(x/y^2)^2}=\frac{z}{1+z^2}\end{align}$, where $z=x/y^2$ (and $y\neq0$).
So for $z$ large enough, our expression will go to 0. Now we only need to worry for the case that $z$ approaches 0. We rewrite again: $\begin{align}f(x,y)=\frac{1}{1/z+z}\end{align}$, so $\lim_{z\to0}\frac{1}{1/z+z}=0$. Whatever $(x,y)$ do, if their values get small enough, we see that $f(x,y)$ will get arbitrarily close to zero, en if their values get big enough, $f(x,y)$ also comes arbitrarily close to zero.
However, this can't be true, because $f(cy^2,y)=\frac{c}{1+c^2}$, so the function isn't even continuous on 0 to begin with. I'm stuck; can someone help me with this?

*

*Consider $(z+1)^2 \geq0$. This leads to $\Large \frac{z}{z^2+1}$$\geq-$$\Large\frac{1}{2}$.

*Now take  $(z-1)^2 \geq0$ This yields $\Large \frac{z}{z^2+1}$$\leq$$\Large\frac{1}{2}$. 
Therefore $\large f $ is bounded.
