We have,
$f(x) =3 x^{3}-9 x^{2}-27 x+15 $
$\Rightarrow f'(x) =9 x^{2}-18 x-27$
For maxima or minima, we put $f^{\prime}(x)=0$
$\Rightarrow 9 x^{2}-18 x-27=0$
$\Rightarrow x^{2}-2 x-3=0$
$\Rightarrow (x-3)(x+1)=0 \Rightarrow x=-1,3$
Now, $f''(x)=18 x-18$
at $x=-1, f''(x)=18(-1)-18=-36 <0$
So, $x=-1,$ point of maxima
$\therefore $ Maximum value of $f(x)$ at $x=-1$ is
$f(-1) =3(-1)^{3}-9(-1)^{2}-27(-1)+15$
$=-3-9+27+15=-12+42=30$