Answer
$9x^{2}+\displaystyle \frac{1}{\sqrt[3]{x^{2}}}$
Work Step by Step
Written in exponential form,
... sum/difference rule...
$f^{\prime}(x)=\displaystyle \frac{d}{dx}[3x^{3}]+\frac{d}{dx}[3x^{1/3}]$
... constant multiple ...
$=3\displaystyle \frac{d}{dx}[x^{3}]+3\frac{d}{dx}[x^{1/3}]$
... power rule...
$=3(3x^{2})+3(\displaystyle \frac{1}{3}x^{-2/3})$
$=9x^{2}+x^{-2/3}$
$=9x^{2}+\displaystyle \frac{1}{\sqrt[3]{x^{2}}}$
