Let $f$ be a real function such that $f(x-y), f(x) f(y)$ and $f(x+y)$ are in A.P. for all $x, y \in R$. If $f(0) \neq 0$, then

$f^{\prime}(3)+f^{\prime}(-3)=0$

It is given that $f(x-y), f(x) f(y)$ and $f(x+y)$ are in A.P. for all $x, y \in R$.

∴ $2 f(x) f(y)=f(x-y)+f(x+y)$ for all $x, y \in R$        ......(i)

Putting x = y = 0, we get

$2\{f(0)\}^2=2 f(0)$

$\Rightarrow f(0)\{f(0)-1\}=0 \Rightarrow f(0)=1$        [∵ f(0) ≠ 0]

Putting x = 0 and y = x in (i), we get

$2 f(0) f(x)=f(-x)+f(x)$ for all $x \in R$

$\Rightarrow 2 f(x)=f(-x)+f(x)$ for all $x \in R$

$\Rightarrow f(-x)=f(x)$ for all $x \in R$

$\Rightarrow f(x)$ is an even function

$\Rightarrow f^{\prime}(x)$ is an odd function

∴  $f(-1)=f(1)$ and $f^{\prime}(-3)=-f^{\prime}(3)$

$\Rightarrow f(1)=f(-1)$ and $f^{\prime}(3)+f^{\prime}(-3)=0$