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.

$$2f(x)f(y)=f(x+y)+f(x-y)$$
Clearly $$f(x)=\cos x$$ satisfies above equation.
