#### Answer

Rolle's Theorem can be applied; $c=\frac{\pi}{6}.$

#### Work Step by Step

$f(x)$ is continuous for all values of $x$ and is differentiable at every value of $x.$
$f(0)=f(\frac{\pi}{3})=0.$
Since $f(x)$ is continuous over $[0 , \frac{\pi}{3}]$ and differentiable over $(0, \frac{\pi}{3})$, applying Rolle's Theorem over the interval $[0, \frac{\pi}{3}]$ guarantees the existence of at least one value $c$ such that $0\lt c\lt \frac{\pi}{3}$ and $f'(c)=0.$
$f'(x)=3\cos{3x}.$
$f'(x)=0\to3\cos{3x}=0\to3x=\frac{\pi}{2}\to c=\frac{\pi}{6}.$