An odd function $f$ is differentiable at zero. Prove $f'(0)=0$? I know that $f'$ of an even function is odd function, thus I have $f(x)=f(-x)$.
However I'd no idea how to prove that $f'(0)=0$?
Please answer my question...

I assume you mean that $f(x)$ is an even function. If so, note that for an even function $f(x)$,  it satisfies $f(x)=f(-x)$. Now if a function is differentiable, it satisfies
$$f'(x) =\lim_{h\to0^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}$$
Now note that
$$f'(0)=\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to0^+}\frac{f(-h)-f(0)}{-h}$$$$=-\lim_{h\to0^+}\frac{f(h)-f(0)}{h}=-f'(0)$$
So we have $f'(0)=-f'(0)\implies f'(0)=0$
