66First we notice that f(x) > 0; for if f(k) = 0 for some k then the setting y = k in the original equation we obtain,
f(x+k) f(x) = f(x) for all x which yields f(x) = 0 or f(x) = 1 for all x. But definitely then the derivative at 0 can't be 2 anywhere.
Letting f(0) = a and setting y = a in the original equation, we get
f(x)2 = f(x+a)
Since f(x) > 0, let g(x) = log f(x) for some base b > 0 and not equal to 1. The above equation then gives
2 g(x) = g(x+a)
(This equation is one of the so called conjugacy equations, the general form being
\varphi(g(x)) = \psi (g(x))
for appropriately chosen functions \varphi and \psi in the unkown g(x))
I am not familiar with the general solution, but for the current situation we can easily obtain
g(x) = g(0) 2x/a
As such the function f is of the form
f(x) = c2x/a with appropriate choice of a being determined from the condition f'(0) = 2
