$f(x+y)=\frac{2f(x)f(y)-f(x)-f(y)}{f(x)f(y)-1}\Rightarrow$ $f(x+y)-1=\frac{f(x)f(y)-f(x)-f(y)+1}{f(x)f(y)-1}\Rightarrow$ $f(x+y)-1=\frac{(f(x)-1)(f(y)-1)}{f(x)f(y)-1}\Rightarrow$ let $g(x)=f(x)-1$ we have $g(x)$ is continuous and $g(x+y)=\frac{g(x)g(y)}{g(x)g(y)+g(x)+g(y)}\Rightarrow$ $\frac{1}{g(x+y)}=\frac{g(x)g(y)+g(x)+g(y)}{g(x)g(y)}\Rightarrow$ $\frac{1}{g(x+y)}=\frac{g(x)+g(y)}{g(x)g(y)}+1\Rightarrow$ $\frac{1}{g(x+y)}=\frac{1}{g(y)}+\frac{1}{g(x)}+1\Rightarrow$ now let $h(x)=\frac{1}{g(x)}+1$ so we have $h(x)$ is additive and continuous so we are done.

D_Phoenix wrote: $f(x+y)=\frac{2f(x)f(y)-f(x)-f(y)}{f(x)f(y)-1}\Rightarrow$ $f(x+y)-1=\frac{f(x)f(y)-f(x)-f(y)+1}{f(x)f(y)-1}\Rightarrow$ $f(x+y)-1=\frac{(f(x)-1)(f(y)-1)}{f(x)f(y)-1}\Rightarrow$ let $g(x)=f(x)-1$ we have $g(x)$ is continuous and $g(x+y)=\frac{g(x)g(y)}{g(x)g(y)+g(x)+g(y)}\Rightarrow$ $\frac{1}{g(x+y)}=\frac{g(x)g(y)+g(x)+g(y)}{g(x)g(y)}\Rightarrow$ $\frac{1}{g(x+y)}=\frac{g(x)+g(y)}{g(x)g(y)}+1\Rightarrow$ $\frac{1}{g(x+y)}=\frac{1}{g(y)}+\frac{1}{g(x)}+1\Rightarrow$ now let $h(x)=\frac{1}{g(x)}+1$ so we have $h(x)$ is additive and continuous so we are done. You can't divide by $f(x)f(y)-1$ because it might be $0$ for some reals. The same can be said about the substitution $g(x)=f(x)-1$ and having $\frac{1}{g(x)}$. The ending isn't ok either as you haven't checked even the possible solution which might not be the only one by the way. Please review and correct your solution.

oh u r right! sorry