Answer
$f(x)$ is increasing on (-$\infty$,0) $\cup$ (0,$\infty$)
Work Step by Step
$f$($x$) =$x^{3}$
$f'(x)$=$3x^{2}$
$f'(x)=0$ when $x=0$
Therefore, $x=0$ is the critical point of $f(x)$ and the only point at which the function is not increasing.
When $x<0$, $f'(x)>0$
When $x>0$, $f'(x)>0$
Therefore, when x $\ne$ 0, $f(x)$ is increasing.
