Differentiable function such that $f(x+y),f(x)f(y),f(x-y)$ are an arithmetic progression for all $x,y$ If $f$ is a differentiable function on $\mathbb{R}$ such that $f(x+y),f(x)f(y),f(x-y)$(taken in that order) are in arithmetic progression for all $x,y\in \mathbb{R}$ and $f(0)\neq0,$then
$(A)f'(0)=-1\hspace{1cm}(B)f'(0)=1\hspace{1cm}(C)f'(1)-f'(-1)=0\hspace{1cm}(D)f'(1)+f'(-1)=0$
My Try:
As $f(x+y),f(x)f(y),f(x-y)$ are in A.P.
$2f(x)f(y)=f(x+y)+f(x-y)$.........(1)
Put $x=0,y=0$ in the above equation
$2f^2(0)=2(0)$
$f(0)=1$ because $f(0)\neq0$
Differentiating the equation $(1)$,
$2f(x)f'(y)\frac{dy}{dx}+2f(y)f'(x)=f'(x+y)(1+\frac{dy}{dx})+f'(x-y)(1-\frac{dy}{dx})$
I am stuck here .Now i dont know how to go further.Please help me.


$f(x+y) ,f(x)f(y),f(x-y)$ are an arithmetic progression 

All nontrivial twice-differentiable solutions are $f(x) = \frac{e^{Ax} + e^{-Ax}}{2}$ for $A$ real or pure imaginary.
The functional equation is $f(x+y) + f(x-y) = 2f(x)f(y)$.
For small $y$, twice-differentiable solutions satisfy $$2f(x) + y^2 f''(x) + o(y^2) = 2f(x)f(y).$$  If there is any point $x$ where $f''(x)f(x) \neq 0$, that implies $2f(x)(f(y)-1) = O(y^2)$ near $y=0$.  It is then possible to divide by $y^2$ and the limit as $y \to 0$ necessarily exists:
$$ f''(x) = 2f(x) \lim_{y \to 0} \frac{f(y)-1}{y^2} = f(x) f''(0)$$
so that $f''(x) = Cf(x)$ for constant $C$ (and $f(0)=1$ and $f'(0)=0$).
Taking $A^2 = C$ yields the stated solutions.
