Determining all $f : \mathbb R \to \mathbb R$ that satisfy $xf(x) - yf(y) = (x-y)f(x+y)$ 
Determine all $f : \mathbb R \to \mathbb R$ that satisfy
$$xf(x) - yf(y) = (x-y)f(x+y) ; \ \forall x , y \in \mathbb R$$

I tried reform the equation to
$$f(x) + f(-x) = 2f(0)$$
and
$$f(x) + f(3x) = 2f(2x).$$
Also, $f$ is injective if $f$ isn't constant.
Can anyone give me some hints please. Thank you very much!
 A: Write the given functional equation for the three pairs $(x,y)$, $(y,z)$, $(z,x)$, and add up. You then obtain
$$0=(x-y)f(x+y)+(y-z)f(y+z)+(z-x)f(z+x)\ .$$
We now have a functional equation with three free variables. Put 
$$x:={t+1\over2},\quad y:={t-1\over2},\quad z:={1-t\over2}\ ,$$
and you get
$$0=1\cdot f(t)+(t-1)\cdot f(0)-t\cdot f(1)\ ,$$
or
$$f(t)=f(0)+t\bigl(f(1)-f(0)\bigr)\qquad\forall t\in{\mathbb R}\ .$$
This shows that $f$ has to be of the form $f(t)=at+b$ with arbitrary constants $a$, $b$.
A: Here are a few hints : use $\frac{xf(x)-(y+z)\bigg(\frac{yf(y)-zf(z)}{y-z}\bigg)}{x-(y+z)}=\frac{(x+y)\bigg(\frac{xf(x)-yf(y)}{x-y}\bigg)-zf(z)}{(x+y)-z}$ to show that $f(z)=\frac{xf(y)-yf(x)+z(f(x)-f(y))}{x-y}$. Deduce that there are two constants $a,b$ such $f(z)=az+b$, then plug into the original equation.
A: Here's another interesting solution:
$$xf(x)−yf(y)=(x−y)f(x+y)\iff x(f(x+y)-f(x))=y(f(x+y)-f(y))$$
Let,
$$g(x)=\frac{f(x)-f(0)}{f(1)-f(0)} \iff g(x)(f(1)-f(0))+f(0)=f(x)$$
So, now, $g(1)=1$ and $g(0)=0$ and
$$x(g(x+y)-g(x))=y(g(x+y)-g(y))$$
$$y=1 \implies x(g(x+1)-g(x))=g(x+1)-1 \tag{1}$$
$$y=-x \implies-g(x)=g(-x) \tag{since $g(0)=0$}$$
$$y=1-x \implies x(1-g(x))=(1-x)(1-g(1-x))$$
Replacing $x$ with $-x$ in the previous equation yields (knowing the fact that $g$ is odd)
$$ -x(1+g(x))=(x+1)(1-g(x+1))\iff 2x+1=x(g(x+1)-g(x))+g(x+1)$$
$$\implies2g(x+1)=2(x+1) \tag{by (1)}$$
$$\implies g(y)=y  \ \ \ \ \forall  \ y \in \mathbb{R}$$
$$\implies f(x)=ax+b  \ \ \ \ \forall \ a,b,x \in \mathbb{R}$$
