When x = 1, then $f(1)(f(-y)-1) = 0$, so $f(1) = 0$ ( $f(1) \neq 0$ implies $f(x) \equiv 1$, which is not a solution) Now let's substitute $(x,y) = (-1, -1)$: $f(-1)(f(-1) - 2) = 0$. Since function is suriective, for any $x \in R - {1}$ there exist $a$ such that f(a) = x. Substituting a to the second equation $f(x) = \frac{1}{f( \frac{1}{a})}$. Now it's easy to see, that $f(x) = 0 \Leftrightarrow x =1$. Two more subtitutes: $(a, \frac{1}{a}) \Rightarrow 0 = f( \frac{1}{a})f(-a)-f( \frac{1}{a}) + x \Leftrightarrow xf(x) = 1 - f(-a)$ (*) $( \frac{1}{a} , a) \Rightarrow 0 = xf( \frac{-1}{a} ) - x + f( \frac{1}{a}) \Leftrightarrow xf(x) = \frac{1}{1 - f( \frac{-1}{a})}$ (**) Let $A = f(-a)$ and $B = f( \frac{-1}{a})$. Connecting (*) and (**): $(1-A)(1-B) = 1$ (***). One more substitution: $(\frac{-1}{a}, -a) \Rightarrow A=B(1-x)$. This and (***) implies $B(-2+x + B(1-x))=0$. Let's notice, that $B = 0 \Rightarrow a=-1 \Rightarrow x=2$, so $B= \frac{2-x}{1-x}$. Using this in (**) we finally get $f(x)= \frac{x-1}{x}$, which is obviously a solution.

Let P(x, y) : $\color{white}\ . \ \color{black}\ \quad f(xy)=f(x)f(-y)-f(x)+f(y)$ if you consider P(x, y) and P(y, x) and P(-x, - y) you can prove that f(-x) =2-f(x) now we know P(x, y) :f(xy) =-f(x). f(y) +f(x)+f(y) P $(x, \frac{1}{x})$ : $\frac{1}{f(\frac{1}{x})} = \frac{f(x) - 1}{f(x)} $ for any x such that f(x) and $f(\frac{1}{x})$ $\neq 0$ and $x\neq1$ If you consider that $\color{white}\ . \ \color{black}\ \quad f(f(x))=\frac{1}{f(\frac{1}{x})}$ you can prove that f(x) =0 if and only if x=1 now we know that f(f(x)) =$\frac{f(x) - 1}{f(x)}$ Since function is suriective, for any $x \in R - {1}$ we have f(x) = $\frac{x-1}{x}$ and proof is finished now