Let $f$ and $g$ be functions such that $f''$ and $g''$ exist. Let, for some $a, $f(a)=f(b)=g(a)=g(b)=0$, and $g''(x) \neq 0$ for every $x \in (a,b)$. Show that there exists $c \in (a,b)$ such that
$\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$
Let $f$ and $g$ be functions such that $f''$ and $g''$ exist. Let, for some $a, $f(a)=f(b)=g(a)=g(b)=0$, and $g''(x) \neq 0$ for every $x \in (a,b)$. Show that there exists $c \in (a,b)$ such that
$\frac{f(c)}{g(c)}=\frac{f''(c)}{g''(c)}$
Let $h(t)=f(t)g'(t)-g(t)f'(t),$ and apply Rolle's theorem.
(This assumes $g\ne0$ on $(a,b)$ as well.)