Answer
$y'=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}+\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$
Work Step by Step
$y=f(x)+g(x)\rightarrow f(x)=\sin{\sqrt[3]{x}}$; $g(x)=\sqrt[3]{\sin{x}}$
$f'(x)$ is found using the Chain Rule:
$u=\sqrt[3]{x}$; $\dfrac{du}{dx}=\dfrac{1}{3\sqrt[3]{x^2}}$
$f(u)=\sin{u};\dfrac{d}{du}f(u)=\cos{u}$
$\dfrac{d}{dx}f(x)=\dfrac{d}{du}f(u)\times\dfrac{du}{dx}=(\dfrac{1}{3\sqrt[3]{x^2}})(\cos{\sqrt[3]{x}})=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}.$
$g'(x)$ is found using the Chain Rule:
$u=\sin{x}$; $\dfrac{du}{dx}=\cos{x}$
$g(u)=u^{\frac{1}{3}};\dfrac{d}{du}g(u)=\dfrac{1}{3\sqrt[3]{u^2}}$
$\dfrac{d}{dx}g(x)=\dfrac{d}{du}g(u)\times\dfrac{du}{dx}=(\cos{x})(\dfrac{1}{3\sqrt[3]{sin^2{x}}})=\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$
$y'=f'(x)+g'(x)=\dfrac{\cos{\sqrt[3]{x}}}{3\sqrt[3]{x^2}}+\dfrac{\cos{x}}{3\sqrt[3]{sin^2{x}}}.$
