Is Lagrange's mean value theorem is valid on $x^{1/3}$ in $[-1,1]$? Suppose I have a function $f(x)=x^{\frac{1}{3}}$ in interval $[-1,1]$ , is lagrange's mean value theorem valid here ? $f(x)$ is continuous in this interval but there's a confusion in it's derivative. What about it's derivative at $x=0$ it looks like not defined as $f'(x)=\frac{1}{3}x^{-\frac{2}{3}}$.
Lagrange's mean value theorem in it's standard form can not be applied since as you correctly point out $x^{\frac{1}{3}}$ does not have a finite derivative at $0$. To be perfectly honest off the top of my head I can't think of a function which has the behaviour of $x^\frac{1}{3}$  (i.e. has an infinite derivative at one point and is differentiable everywhere else) yet does not satisfy the conclusion of the mean value theorem. Which should not be taken to mean there are no such functions though.
The theorem requires the function to be differentiable in $(-1,1)$, and it isn't so you can't use it. However, there is a point where $f'(c)=\frac{f(b)-f(a)}{b-a}$:
$$ f'(c)=\frac{1}{3 \sqrt[3]{c^2}} = \frac{f(1)-f(-1)}{1-(-1)} = 1 \Leftrightarrow \sqrt[3]{c^2}=\frac{1}{3} \Leftrightarrow c=\pm \sqrt{\frac{1}{27}}$$