Answer
f is decreasing in $(-\infty, \frac{-5}{3}) \cup (3, +\infty) $ and f is increasing in $(\frac{-5}{3},3)$
Work Step by Step
$f(x)=-x^{3}+2x^{2}+15x+16$
$f'(x)=-3x^{2}+4x+15$
$f'(x)=0 \rightarrow -3x^{2}+4x+15=0 \rightarrow x=3, x=\frac{-5}{3}$
$x \lt \frac{-5}{3} \rightarrow $ f is decreasing
$\frac{-5}{3} \lt x \lt 3 \rightarrow $ f is increasing
$x \gt 3 \rightarrow $f is decreasing