Let $f$ be a continuous and differentiable function such that $f(a)=f(b)=0$ , show that $f'(c)=\pi f(c)$ for some $c \in (a,b)$

Define $g(x)=f(x)\cdot e^{-\pi x}$. Notice that $g$ is, like $f$, continuous on $[a,b]$, differentiable on $(a,b)$, and zero at the endpoints $a$ and $b$. Also notice that $$g'(x)=f'(x)\cdot e^{-\pi x}+f(x)\cdot(-\pi)e^{-\pi x}= (f'(x)-\pi f(x))e^{-\pi x}.$$ Apply Rolle's theorem.

Here's the crux of the matter (I think). Suppose $f>0$ on $(a,b).$ Then $\ln f$ is well defined and differentiable on $(a,b).$ We have $\ln f(x) \to -\infty$ at $a^+,b^-.$ (Good to draw a picture.) It follows from the MVT that $(\ln f)'$ takes on arbitrarily large positive and negative values. By Darboux, $(\ln f)'$ takes on all values in $\mathbb {R},$ in particular the value $\pi.$ Since $(\ln f)' = f'/f,$ we're done.

Now we don't have $f>0$ on $(a,b)$ probably. So the argument is not complete, but I think this is the idea.