how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and 
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then 
$f(x)=f\left( \frac{x+1}{2}  \right)+f\left( \frac{x-1}{2}  \right)$.
But how to find all functions?

We can prove that if a function $f:\mathbb R\to\mathbb R$ satisfies
$$f\left(x^2-y^2\right)=(x-y)\big(f(x)+f(y)\big)\tag0\label0$$
for every real numbers $x$ and $y$, then there is a constant real number $k$ such that $f(x)=kx$ for every real number $x$.
First, let $y=x$ in \eqref{0} and get $f(0)=0$. Next, let $x=1$ and $y=-1$ in \eqref{0} and get $f(-1)=-f(1)$. Now, define $k:=f(1)$ and $g(x):=f(x)-kx$. So we have $g(1)=g(-1)=0$ and:
$$g\left(x^2-y^2\right)+k\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)+k(x+y)\big)$$
$$\therefore\quad g\left(x^2-y^2\right)=(x-y)\big(g(x)+g(y)\big)\tag1\label1$$
Thus letting $y=\pm1$ in \eqref{1} we have:
$$g\left(x^2-1\right)=(x\pm1)g(x)$$
Hence, subtracting these two equations, we conclude that $g(x)=0$ for every real number $x$, which shows that $f(x)=kx$ for every $x$.
