If f(x - y), f(x) f(y) and f(x + y) are in A.P. for all x, y ∈ R and f(0) = 0, then

f'(-2) = f'(2)

It is given that for all $x, y \in R$

f(x - y), f(x) f(y)  and  f(x+y) are in A.P.

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

$\Rightarrow 2 f(0) f(x)=f(-x)+f(x)$ for all $x \in R$            [Replacing x by 0 and y by x]

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

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

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

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

$\Rightarrow f'(-2)=f'(2)$