Consider a function $g(x)$ such that $g'(x) = \frac{1}{a-x} + \frac{1}{b-x} + \frac{1}{a+b}$. Integrating, $g(x) = \frac{1}{(a-x)^2} + \frac{1}{(b-x)^2} + \frac{x}{a+b} + C$. Now consider the function $h(x) = f(x) - g(x)$. We wish to show $\exists c : h'(c) = 0$. Now, $h(x)$ is continuous and differentiable on $(a, b)$, but $\lim_{x \to a} h(x) = \lim_{x \to b} h(x) = - \infty$. Intuitively, there must be some $h'(c) = 0$... a limit with the Mean Value Theorem?

i don't know, my intuition tells me $f$ (of $c$ or of $a$ or something) must appear somewhere in the relation, not just $f'$... but i could be wrong, i have no idea...for example, we can get one possible relation by doing the following. let $g(x) = (a+b)(a-x)(b-x)f(x)+(a-x)(b-x)x$. since $g(a) = g(b) = 0$, by the MVT there's $c\in (a,b)$ s.t. $0 = (a+b)(c-a)f(c)+(a+b)(c-b)f(c)+(a-c)(b-c)(a+b)f'(c)+(a-c)(b-c)+c(c-a)+c(c-b)$, from which we get $f'(c) = \frac{f(c)}{a-c}+\frac{f(c)}{b-c}+\frac{1}{a+b}+\frac{c(a-c)+c(b-c)}{(a-c)(b-c)(a+b)}$ which, yes, isn't very promising if the relation they ask us to prove is actually true.

We consider the function $g(x) =(x-a)(x-b) \, e^{f(x)-\frac{x}{a+b}}$ which is continuous on $[a,b]$ and differentiable in $(a,b)$. It is $g(a)=g(b)=0$ therefore by Rolle's theorem there exists a $c \in (a,b)$ such that $g'(c)=0$. But $g'(x)=[(x-b) + (x-a) + (f'(x)-\frac{1}{a+b})(x-a)(x-b)] \, e^{f(x)-\frac{x}{a+b}}$ so $g'(c)=0 \Longrightarrow (c-b) + (c-a) + (f'(c)-\frac{1}{a+b})(c-a)(c-b)=0$ or $f'(c) =\frac1{a-c}+\frac1{b-c}+\frac1{a+b}$ Q.E.D.

yes, yes, oh goodness, i love it.