A functional equation defined on the real axis. $$
f: \mathbb{R} \to \mathbb{R}\qquad \frac{f(x+y)}{x+y} = \frac{f(x)-f(y)}{x-y}, \qquad \forall x,y\in \mathbb{R}, \left|x\right| \neq \left|y\right|
$$
Can I prove anything interesting about this function? I need to find it.
 A: Since $|x|\not = |y|$, I'll let $x\notin \{-\frac{1}{2}, 0, \frac{1}{2}\}$ for the following substitutions
Observe first that, for $x\to -x$ and $y=x+1$ $$\frac{f(1)}{1}=-\frac{f(-x)-f(x+1)}{2x+1}\iff f(-x)-f(x+1)=-(2x+1)·f(1)\tag{I}$$
Letting $x\to x+1\text{ and }y=x-1$  we obtain $$\frac{f(2x)}{2x}=\frac{f(x+1)-f(x-1)}{2}\iff f(x+1)-f(x-1)=\frac{f(2x)}{x}\tag{2}$$
Finally $x\to x-1$ and $y\to-x$: $$\frac{f(-1)}{-1}=\frac{f(x-1)-f(-x)}{2x-1}\iff f(x-1)-f(-x)=-f(-1)·(2x-1)\tag{3}$$
The trick now is to add $(1), (2)$ and $(3)$ since the LHS turns $0$

$$\begin{array}a &0&=-(2x+1)·f(1)+\frac{f(2x)}{x}-f(-1)·(2x-1)\\ \iff&0&=-2x^2·f(1)-x·f(1)+f(2x)-2x^2·f(-1)+x·f(-1)\\ \iff &-f(2x)&=-2x^2·\big(f(1)+f(-1)\big)-x·\big(f(1)-f(-1)\big)\\ \iff& f(x)& =\frac{1}{2}·x^2·\big(f(1)+f(-1)\big)+\frac{1}{2}·x·\big(f(1)-f(-1)\big) \end{array}$$

Which is basically a quadratic equation in $x$. Can you end it now?
A: If $y\to x$ then
$$\dfrac{f(2x)}{2x} = f'(x),$$
$$f(2x) = 2x f'(x)\tag1.$$
Let Maclaurin series of $f(x)$ are
$$f(x) = c_0 + c_1x+c_2x^2+c_3x^3+c_4x^4+\dots,\tag2$$
then $(1)$ leads to the identity
$$c_0 + 2c_1x+4c_2x^2+8c_3x^3+16c_4x^4+\dots = 2x(c_1 + 2c_2x+3c_3x^2+4c_4x^3+\dots),$$
which satisfies iff 
$$c_0=c_3=c_4=\dots = 0.$$
Easy to check that the function
$$f(x)=c_1x+c_2x^2\tag3$$
satisfies to the given equation for $(c_1,c_2)\in\mathbb R^2$.
The Laurent series (with the negative degrees of $x$) leads to the same result.
A: See Aczél or Schwaiger, where there is a misprint: $h(x_1,\ldots,x_n)$ should read as $$h(x_1+\ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $\frac{f(x)-f(y)}{x-y}=g(x+y)$ for $x\not=y$ and certain generalisations.
