This is how I have the proof for this set up. However now I'm not really sure that saying "f'(x)=(f(1)-0)/1=f(1). This is equivalent to f'(x)=f(b). Hence we can write that f'(x)≥f(x)." is the right route to go? Does this look like the right direction?
Hint: Use the fact that $f$ attains it maximum on the closed interval $[0,\frac12]$ at some $x_0\in [0,\frac12]$. Assume $f(x_0)>0$ and invoke the IVT to find $0<\xi<x_0$ with $f'(\xi)=\frac{f(x_0)-f(0)}{x_0-0}> 2 f(x_0)$ and use that $2f(\xi)\ge f'(\xi)$. From this you obtgain a contradiction and hence deduce that $f(x)=0$ for $0\le x\le\frac13$. How can you proceed to the rest of the interval?
Okay looking at the theorems and some stuff similar this is what I am coming up with now. I feel this might make more sense of the proof? 
