Functional equation $ f ( x + y ) f ( x - y ) = f ^ 2 ( x ) $ 
Let a function $ f $ be continuous and differentiable for all $ x $, such that it satisfies
$$ f ( x + y ) f( x - y ) = f ^ 2 ( x ) \text . $$
Given that $ f ( 0 ) $ is nonzero and $ f ( 1 ) $ is $ 1 $, find $ f $.

I tried replacing $ x $ by $ y $ and then by $ - x $, but was not able to proceed further.
 A: Hints: the answer is $f(x)=e^{a(x-1)}$ for some conatant $a$. First prove that $f(t)=0$ for some $t$ leads to $f \equiv 0$, a contradiction. Then prove that $f(x) >0$ for all $x$. Let $g(x)=\ln\, f(x)$ and show that $g(\frac {a+b} 2)= \frac {g(a)+g(b)} 2$. From this we get $g(at+(1-a)s)=ag(t)+(1-a)g(s)$ for  dyadic rational $a$, hence for all $a$ (by continuity). Conclude that $g(x)$ has the form $ax+b$
A: With $x=y$, $f(2x)f(0)=f^2(x)$ and $f$ has a constant sign and is positive ($f(1)=1$). 
Let $$h(x):=\log\frac{f(x)}{f(0)}.$$
Setting $x=y$, we have
$$h(2x)=2h(x),$$
then with $u:=x+y,v:=x-y$,
$$h(u)+h(v)=2h\left(\frac{u+v}2\right)=h(u+v)$$
so that $h$ must be linear.
After a little computation,
$$f(x)=f_0^{1-x}.$$
A: Hint.
Assuming $f(0) \ne 0$ we have that $f(y)f(-y) = f^2(0)$ then we conclude that $f(0) \ne 0 \to f(x) > 0\to f(0) > 0$
so making
$$
\log_u f(x+y)+\log_u f(x-y) = 2\log_u f(x)
$$
or
$$
g(x+y) + g(x-y) = 2g(x)
$$
and now making $g(x) = a x + b$
$$
a(x+y)+b +a(x-y) + b = 2a x + 2 b
$$
we have a good estimation as
$$
\log_u f(x) = a x + b\to f(x) = u^{ax+b}
$$
Of course if $f(0) = 0$ then $f(x) = 0$ is also a solution.
