An alternative method:

Given that f(x) = f(y) f(x-y)

Put x = y = 0

we get f(0) = f²(0). So, f(0) = 1 or f(0) = 0

But if f(0) = 0, we get f(x) = 0 for all x, and in any case it contradicts  as f'(0) and p are both real numbers.

Hence f(0) = 1

Now let

Since f is differentiable everywhere,