Use the product rule: the derivative of f(x)f(-x) is f'(x)f(-x)+ f(x)(f(-x))' which, by the chain rule is f'(x)f(-x)- f(x)f'(-x). Since we have that f'(x)= f(x), we also have that f'(-x)= f(-x) and so (f(x)f(-x))'= f(x)f(-x)- f(x)f(-x)= 0.

(You don't need f(0)= 1. This is true for any exponential . which is a constant and has derivative 0.)