$f(xy)=\frac{f(x)+f(y)}{x+y}$ 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.

Let $y=1$
$f(x)(x+1)=f(x)+f(1)$ Hence $$f(x)=\frac{f(1)}{x}$$
You can verify that : $$f(xy)=\frac{f(1)}{xy}=\frac{\frac{f(1)}{x}+\frac{f(1)}{y}}{x+y}$$
As Peter Taylor explained in the comment :
$$f(0)=f(1)+f(0)$$
So if you want $f$ to be defined on $\mathbb R$, you need to have $f(1)=0$. Hence $$\forall x\neq 0\quad f(x)=\frac{f(1)}{x}=0$$
Then $\forall x\neq 0\quad x.f(0)=f(0)$ implies $f(0)=0$
