$f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ For all $x,y\in\mathbb{R}$. also $f : \mathbb{R} → \mathbb{R}$ and $x+y\not=0$.
My attempt:
 I restated it as 
$a[x^2 y^2 (\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2})] +
 b[xy(x+y-\frac{1}{y}-\frac{1}{x})] + c [x+y-2]=0$
because of $f(xy) - \frac{f(x)+f(y)}{x+y}=0$
(we know that $f$ is identically equal to $0$)
and later tried to prove that 
$\frac{x}{y}+\frac{y}{x}-\frac{1}{y^2}-\frac{1}{x^2}$
$x+y-\frac{1}{y}-\frac{1}{x}$
$x+y-2$
are all equal to 1. I eventually got to that 
$x^2 -2x -1,5 =0$ And i don't like this
Later I tried ;
$\frac{a(x^2+y^2)+b(x+y)+2c}{x+y} = \frac{f(x+y)}{x+y}$
because if
$(x+y)^2=x^2+y^2$
$xy=0$ but that implies that $c$ is equal to $0$ too. Both of my trials are propably wrong.

Setting $(x, y) = (1, 0)$ gives $f(1) = f(1) + f(0)$; that is, $f(0) = 0$. For $x\not =0$, setting $y = 0$ in the given equation shows that $(f(x) + f(0))/x = f(x)/x$ vanishes; that is, $f(x) = 0$.
