If the solve the problem

Given : $f(x)=x^{3}-3 x$

$\Rightarrow f^{\prime}(x)=3 x^{2}-3$

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow 3 x^{2}-3=0$

$\Rightarrow x^{2}-1=0$

$\Rightarrow x=\pm 1$

Since $f^{\prime}(x)$ changes from negative to positive as $x$ increases through $1, x=1$ is the point of local minima. The local minimum value of $f(x)$ at $x=1$ is given by

$(1)^{3}-3(1)=-2$

Since $f^{\prime}(x)$ changes from positive to negative when $x$ increases through $-1, x=-1$ is the point of local maxima.

The local maximum value of $f(x)$ at $x=-1$ is given by

$(-1)^{3}-3(-1)=2$