Identifying why the MVT does not apply

Differentiating, $f'(x) = \frac{2}{3}x^{-1/3} = \dfrac{2}{3\sqrt[3]{x}}$. This expression is undefined at $x = 0$, since it involves dividing by $x^{1/3}$, which equals $0$ there. Note that $f(x) = x^{2/3}$ itself is continuous everywhere, including at $x=0$, so continuity on $[-1,1]$ is not the issue.

Since $0$ lies in the open interval $(-1,1)$, the function fails to be differentiable at an interior point of the interval. This violates the hypothesis of the MVT that $f$ must be differentiable on the entire open interval $(a,b)$. Because this hypothesis fails, the theorem does not apply, and there is no guarantee that some $c \in (-1,1)$ satisfies $f'(c) = \frac{f(1)-f(-1)}{1-(-1)}$ — indeed, no such $c$ exists here since $f'(x) \neq 0$ for any $x \neq 0$ while $\frac{f(1)-f(-1)}{2}=0$.