The function f(x)=x^3-3 x, is

We have,

$f(x)=x^3-3 x \Rightarrow f'(x)=3 x^2-3$

For f(x) to be increasing, we must have

$f'(x) \geq 0 \Rightarrow 3 x^2-3 \geq 0 \Rightarrow x^2-1 \geq 0 \Rightarrow x \leq-1$ or $x \geq 1$

Hence, f(x) is increasing on $(-\infty,-1] \cup[1, \infty)$ and decreasing on (-1, 1).