Answer
$g'(3) = -\frac{\sqrt{2}}{6}$
Work Step by Step
$g(x) = \sqrt{f(x)}$
$g'(x) = \frac{1}{2}(f(x))^{-1/2}\cdot f'(x)$
$g'(x) = \frac{f'(x)}{2\sqrt{f(x)}}$
$g'(3) = \frac{f'(3)}{2\sqrt{f(3)}}$
$g'(3) = \frac{-\frac{2}{3}}{2\sqrt{2}}$
$g'(3) = \frac{-1}{3\sqrt{2}}$
$g'(3) = -\frac{\sqrt{2}}{6}$
