Derivative of a function is odd prove the function is even.
Let $g(x)=f(-x)$. Then $g'(x)=-f'(-x)=f'(x)$.
Since $g(0)=f(0)$ and $g'=f'$, it follows from the mean value theorem that $g=f$.
Define functions $f_0(x)=(f(x)+f(-x))/2$ and $f_1(x)=(f(x)-f(-x))/2$. Then $f_0$ and $f_1$ are also differentiable, and $f_0$ is even and $f_1$ is odd.
Show that the derivative of an odd function is even, and that of an even function is odd.
From the equality $f'=f_0'+f_1'$ conclude that $f_1$ is constant and, therefore, zero.