= Solution
Set
$$
H(x)=(g(b)-g(a))f(x)-(f(b)-f(a))g(x).
$$
Its endpoint values are equal. <Rolle theorem> therefore supplies $\xi\in(a,b)$ with
$$
(g(b)-g(a))f'(\xi)
=(f(b)-f(a))g'(\xi). \qquad (1)
$$
If $g'(\xi)=0$, then (1) and $g(b)\ne g(a)$ force $f'(\xi)=0$, contrary to the hypothesis. Hence $g'(\xi)\ne0$, and division in (1) gives
$$
\boxed{
\frac{f(b)-f(a)}{g(b)-g(a)}
=\frac{f'(\xi)}{g'(\xi)}}.
$$
The condition is necessary. On $[0,2\pi]$, let
$$
f(x)=\sin x,\qquad
g(x)=\sin x+\frac{x}{4}+\frac{\sin2x}{8}.
$$
Here $g(2\pi)-g(0)=\pi/2$ while $f(2\pi)-f(0)=0$, so the endpoint ratio is zero. But
$$
f'(x)=\cos x,\qquad
g'(x)=\cos x\left(1+\frac12\cos x\right).
$$
Where $g'(x)\ne0$, their ratio is $1/(1+\tfrac12\cos x)$ and is never zero. At the remaining points both derivatives vanish, so the derivative ratio is undefined. Thus the conclusion fails when simultaneous zeros are allowed.
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