Differentiable function such that $f(x+y),f(x)f(y),f(x-y)$ are an arithmetic progression for all $x,y$

Put $y=0$ to obtain $2f(0)f(x)=2f(x)$ and $x=0$ to get $2f(0)f(y)=f(y)+f(-y)$. Thus we get $2f(x)=2f(0)f(x)=f(x)+f(-x)\implies f(x)=f(-x)$.

Since therefore $f$ is even, it has to be answer (D), because: $$ f'(-1)=\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to1}\frac{f(-x)-f(-1)}{-x+1}=-\lim_{x\to1}\frac{f(x)-f(1)}{x-1}=-f'(1) $$

For $x=y$, you have $$f(x-x)+f(x+x)=2f(x)f(x)$$ or $$f(0)+f(2x)=2f^2(x)$$ or, $$2f^2(x)-f(2x)=1$$

Differentiating both sides with respect to $x$, we get $$4f(x)f'(x)-2f'(2x) = 0 $$

Putting $x=0$, you have $$f'(0)=0$$

So options (a) and (b) are wrong.

As for (c) and (d), they are trying to check if $f'(x)$ is an even function or odd.

Use the fact that $$f(2x)=2f^2(x)-1$$ and $$f(-2x)=2f^2(-x)-1$$

Now assuming the function is even, i.e.$$f(2x)=f(-2x) \Rightarrow f^2(x)=f^2(-x) \Rightarrow f(x)=f(-x)$$ That is, the equation supports that the function is even.

But assuming the function is odd, $$f(2x)=-f(-2x)$$ $$\Rightarrow 2f^2(x)-1=1-2f^2(-x)$$ $$\Rightarrow 2[f^2(x)+f^2(-x)]=2$$ $$\Rightarrow [f^2(x)+f^2(-x)]=1 \not \Rightarrow f(x)=-f(-x)$$ That is, the equation does not support that the function is odd.

So your answer is (d).