Ok not to be a pain but the word is differentiable not derivable.
Taking the square root
f ' ^2 -f = 1 or -1
Differentiate -- in either case
2f " f ' -f ' = 0
f ' (2f"-1) = 0
Now solve f ' = 0 (use original equation to find values of c) and 2f " - 1 = 0
That should be enough to get you to the end