Answer
There is a number $c$ in the interval $(-b,b)$ such that $f'(c) = \frac{f(b)}{b}$
Work Step by Step
Since $f$ is differentiable everywhere then $f$ is continuous for all $x$.
We can choose any positive number $b$.
Note that $f(-b) = -f(b)$ since $f$ is an odd function.
According to the Mean Value Theorem, there is a number $c$ in the interval $(-b,b)$ such that $f'(c) = \frac{f(b)-f(-b)}{b-(-b)}$
Then:
$f'(c) = \frac{f(b)-f(-b)}{b-(-b)}$
$f'(c) = \frac{f(b)-(-f(b))}{b+b}$
$f'(c) = \frac{f(b)+f(b)}{b+b}$
$f'(c) = \frac{2f(b)}{2b}$
$f'(c) = \frac{f(b)}{b}$
