real analysis - Let $f(x)$ be a differentiable function at $x=0$ & $fleft(frac{x+y}{k}right)=frac{f(x)+f(y)}k left(kin R, kne 0,2right)$. Show following

Let $f(x)$ be a derivable function at $x=0$ & $f\left(\dfrac{x+y}{k}\right)=\dfrac{f(x)+f(y)}{k} \left(k\in R, k\ne 0,2\right)$. Show that $f(x)$ is either a zero or odd linear function.

My attempt is as follows:-

Putting $x=0,y=0$ in the functional equation

$$f(0)=\dfrac{2f(0)}{k}$$
$$f(0)(k-2)=0$$
$$f(0)=0 \text { as it is given $k \ne 2$ }$$

Replace $y=-x$ in the functional equation.

$$f(0)=\dfrac{f(x)+f(-x)}{k}$$
$$f(x)+f(-x)=0$$

$$f(-x)=-f(x)$$

Thus $f(x)$ is definitely an odd function.

As $f(x)$ is a derivable function at $x=0$ then following limit should exist:-

$$\lim_{h\to 0}\dfrac{f(h)-f(0)}{h}$$
$$\lim_{h\to 0}\dfrac{f(h)}{h}$$

Assuming $f(x)$ to be polynomial function.

If $f(x)$ is a linear function, then limit should be non-zero because if its a polynomial of degree greater than $1$, then limit would be zero.

Like if $f(x)=ax^2+bx+c$

Applying L' Hospital rule:-
$$\lim_{h\to 0}(2ah+b)=0$$

If $f(x)$ is a constant function other than $0$, then $f(x)=-f(x)$ will not hold.

So we just have to prove that $$\lim_{h\to 0}\dfrac{f(h)}{h} \ne 0$$

Applying L' Hospital rule as we have
$\dfrac{0}{0}$ form and function is derivable at $x=0$

$$\lim_{h\to 0}f'(h)=f'(0)$$

If we can just prove that $f'(0) \ne 0$, then we will be done.

I am not getting how to prove this fact.Any hints?