Answer
$x=\pm 1,$
relative minimum at $(1,-1),\quad$relative maximum at $(-1,7).$
Work Step by Step
Critical points of $f$ are points that are either stationary or singular
(for which either $f'(c)=0$ or $f'(c)$ is undefined, but $f(c)$ is)
$g(x)=2x^{3}-6x+3$
$g'(x)=6x^{2}-6\qquad $defined for all x.
$g'(x)=0$
$6x^{2}-6=0$
$6x^{2}=6$
$x^{2}=1$
$ x=\pm 1\qquad$ ... critical points: $(1,-1),(-1,7)$
Second-derivative test:
$g''(x)=6x$
$ g''(-1)=-6 \lt 0$, so f has a relative maximum at $(-1,7)$
$ g''(1)=6 \gt 0$, so f has a relative minimum at $(1,-1)$
