Is a continuous function bounded by Lipschitz continuous function also Lipschitz continuous? 
Is a continuous function bounded by Lipschitz continuous function also Lipschitz continuous?

For example this function
$$f(x,y)=\frac{x^3-3xy^2}{x^2+y^2}$$
and $f(0,0)=0$. I can show that
$$\left|f(x,y)\right|
=\left|\frac{x^3}{x^2+y^2}-\frac{3xy^2}{x^2+y^2}\right|
\leq\left|\frac{x^3}{x^2}\right|+\left|\frac{3xy^2}{y^2}\right|
=4\left|x\right|$$
to prove $f(x,y)$ is continuous. However, since I know $4|x|$ is Lipschitz continuous. Can I also show that $f(x,y)$ is also Lipschitz continuous? Thank you.

No, this doesn't work as stated.
$$f(x)=\begin{cases}0,\quad x<0\\1,\quad x\geq0\end{cases}\text{,}$$
the Heaviside step function, is trivially bounded by $1$ but not even continuous, let alone Lipschitz continuous.
