Checking MVT hypotheses with a vertical tangent

Differentiating gives $f'(x) = \frac{1}{3}x^{-2/3} = \dfrac{1}{3x^{2/3}}$. This expression is undefined at $x=0$ because it involves division by zero, even though $f(0)=0$ is perfectly well-defined. Graphically, this corresponds to a vertical tangent line at $x=0$, since $f'(x) \to \infty$ as $x \to 0$.

Since $0$ lies inside the open interval $(-1,1)$, $f$ fails to be differentiable at every point of $(-1,1)$, so one of the two hypotheses of the MVT (differentiability on the open interval) is not satisfied. Therefore the MVT does not guarantee the existence of such a point $c$ — the theorem simply does not apply here (it may or may not still be true, but MVT gives no guarantee).