$f=f'+f''$ implies $f\equiv 0.$ 
I have trouble understanding why is $f''(d)<0$ in case 1 and $f''(d)>0$ in case 2. I'll appreciate any help.
 A: You are on to something. That's a pretty bad proof. I would argue this way: At $d$ we have $f(d)>0,f'(d)=0.$ From the given equation it follows that $f''(d)=f(d) > 0.$ But if $f''(d) > 0,$ then
$$\frac{f'(d+h)-f'(d)}{h} = \frac{f'(d+h)}{h} > 0$$
for small positive $h.$ This implies $f'>0$ on some interval $(d,d+h_0).$ That in turn implies $f$ is strictly increasing on $[d,d+h_0).$ Thus $f>f(d)$ in $(d,d+h_0).$ That contradicts the fact that $f(d)$ was the maximum value of $f$ on $[0,1].$
A: Your proof is almost fine. Replace in Case 1 the clause 
"Then $f(d)>0$ and $f'(d)=0$ and $f''(d)<0$"
by 
"Then $f(d)>0$,  $f'(d)=0$, and $f''(d)\leq0$, or $f$ would not be locally maximal at $d$."
Similarly in Case 2.
A: If $f(d)$ is a local maximum, you know that the derivative $f'(d) = 0$, and the derivative shortly before $d$ is positive and shortly after $d$, negative. This can only happen if the derivative of the derivative, $f''(d)$ is non-positive. And it can't be 0, because then $f(d)=f'(d)+f''(d)$ would be 0 as well. So it must be negative.
A similar reasoning holds for the second case.
