Answer
See the proof.
Work Step by Step
Using the mean value theorem there exist $c\in (-b,b)$ such that:
$$f'(c)=\frac{f(b)-f(-b)}{b-(-b)}$$
Since $f$ is odd it follows that $f(-b)=-f(b)$ so:
$$f'(c)=\frac{f(b)+f(b)}{b-(-b)}=\frac{f(b)}{b}$$