$f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$

Question

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.

Answer 1

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 $f(0)=0$

Answer 2

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, $f(x) = 0$.