Answer
Rolle's Theorem can be applied $c=\dfrac{\sqrt{3}}{9}.$
Work Step by Step
$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$
$f(0)=f(1)=0.$
Since $f(x)$ is continuous over $[0 , 1]$ and differentiable over $(0, 1)$, applying Rolle's Theorem over the interval $[0, 1]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt 1$ and $f'(c)=0.$
$f'(x)=1-\dfrac{1}{3\sqrt[3]{x^2}}=\dfrac{3\sqrt[3]{x^2}-1}{3\sqrt[3]{x^2}}$.
$f'(x)=0\to \dfrac{3\sqrt[3]{x^2}-1}{3\sqrt[3]{x^2}}=0\to 3\sqrt[3]{x^2}-1=0\to c=\dfrac{\sqrt{3}}{9}.$
