Answer
$y'=\dfrac{4x}{\sqrt[3]{(6x^2+1)^2}}$
Work Step by Step
$u=(1+6x^2)$; $\dfrac{du}{dx}=(12x)$
$y=\sqrt[3]{u};\dfrac{dy}{du}=\dfrac{1}{3\sqrt[3]{u^2}}$
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times\dfrac{du}{dx}=(12x)(\dfrac{1}{3\sqrt[3]{(6x^2+1)^2}})=\dfrac{4x}{\sqrt[3]{(6x^2+1)^2}}$
