Assume that $f$ is differentiable on $(a,b)$ and continuous on $[a,b]$, $f'(x)\neq0$.
Obviously it's related to the mean value theorem, so I was thinking Let $g(x)=e^xf(x)$ and applying the mean value theorem, but that didn't work out well--does the $f'(x)\neq0$ mean that strict monotonicity will be needed to prove this? Or is it just because of the denominator in the question? Any halp would be appreciated! Thanks！
By the MVT, there exists some $c \in (a, b)$ such that $$f'(c) = \frac{f(b) - f(a)}{b - a}.$$ Also, consider $g(x) = f(\ln(x))$, on the interval $[e^a, e^b]$. Then, also by the MVT, there exists some $d$ such that $e^d \in (e^a, e^b)$ and $$e^{-d}f'(d) = \frac{1}{e^d} f'(\ln(e^d)) = \frac{f(\ln(e^b)) - f(\ln(e^a))}{e^b-e^a} = \frac{f(b) - f(a)}{e^b - e^a}.$$ Taking the quotient yields the desired result.
We recognize that the term $\frac{e^b - e^a}{b-a}$ resembles the form given by the Mean Value Theorem, namely $f'(c) = \frac{f(b) - f(a)}{b-a}$ for $a < c < b$.
Let $f(x) = e^x$ and note that $f$ is differentiable on $(a, b)$, continuous on $[a, b]$ and $f'(x) \ne 0$. Then by the Mean Value Theorem, there exists $c$ in $(a,b)$ such that
\begin{align*} f'(c) &= \frac{f(b) - f(a)}{b-a} \\ &= \frac{e^b - e^a}{b-a} \end{align*}
Dividing both sides by $f'(d) = e^d$, we obtain
\begin{align*} \frac{f'(c)}{f'(d)} &= \frac{e^b - e^a}{b-a} \cdot \frac{1}{f'(d)} \\ &= \frac{e^b - e^a}{b-a} \cdot e^{-d} \end{align*}