Answer
$f'(t)=\dfrac{6}{\sqrt[3]{9t+2}}$
Work Step by Step
$u=9t+2$; $\dfrac{du}{dt}=9$
$f(u)=u^{\frac{2}{3}};\dfrac{d}{du}f(u)=\dfrac{2}{3\sqrt[3]{u}}$
$\dfrac{d}{dt}f(t)=\dfrac{d}{du}f(u)\times\dfrac{du}{dt}=\dfrac{6}{\sqrt[3]{u}}=\dfrac{6}{\sqrt[3]{9t+2}}$
