$f'(x)>0,$ so$f(a)<f(b).$ With Lagrange Mean Value Theorem, we know there exists $\xi_1\in(f(a),f(b))\subseteq(a,b),$ such that$$f(f(b))-f(f(a))=f'(\xi_1)(f(b)-f(a)).$$and $\xi_2\in(a,b),$ such that$$f(b)-f(a)=f'(\xi_2)(b-a).$$Hence, we get$$f(f(b))-f(f(a))=f'(\xi_1)f'(\xi_2)(b-a),$$where $f'(\xi_1),f'(\xi_2)>0.$ If $f'(\xi_1)=f'(\xi_2),$ just let $c=\xi_1$ and we get $$f(f(b))-f(f(a))=(f'(c))^2(b-a).$$If$f'(\xi_1)\neq f'(\xi_2),$ then $\sqrt{f'(\xi_1) f'(\xi_2)}$ is between $f'(\xi_1)$ and $f'(\xi_2).$ With Darboux Mean Value Theorem, there exists $c,$ which is between $\xi_1$ and $\xi_2,$ such that $$f(f(b))-f(f(a))=(f'(c))^2(b-a)$$as desired. p.s. It seems that $f'(x)$ does NOT need to be continuous.

Henry_2001 wrote: $f'(x)>0,$ so$f(a)<f(b).$ With Lagrange Mean Value Theorem, we know there exists $\xi_1\in(f(a),f(b))\subseteq(a,b),$ such that$$f(f(b))-f(f(a))=f'(\xi_1)(f(b)-f(a)).$$ i don't think LMVT claims this, even MVT due to Cauchy doesn't imply what you wrote.

Henry_2001 wrote: $f'(x)>0,$ so$f(a)<f(b).$ With Lagrange Mean Value Theorem, we know there exists $\xi_1\in(f(a),f(b))\subseteq(a,b),$ such p.s. It seems that $f'(x)$ does NOT need to be continuous. No. it Is continuous because is $C^1$ and by definition, it MUST to be continuous. Regards Claudio.

look here