Answer
$f'(x)=\sqrt[3] {-\frac{8}{27x^{4}}}$
Work Step by Step
$f(x)=\sqrt[3] \frac{8}{x}=2x^{-\frac{1}{3}}$
$f'(x)=2(-\frac{1}{3})x^{-\frac{1}{3}-1}$
$f'(x)=-\frac{2}{3}x^{-\frac{4}{3}}$
$f'(x)=-\frac{2}{3\sqrt[3] {x^{4}}}$
$f'(x)=\sqrt[3] {-\frac{8}{27x^{4}}}$