$P(1,1): f(1)^2-f(1)-2=0 \implies f(1)=-1$ or $f(1)=2$ $P(x,1): f(x)f(1)=f(x)+\frac{1}{x}+1 \implies f(x) =\frac{\frac{1}{x}+1}{-2}$ or $f(x)=\frac{1}{x}+1$ Thus, checking, we see that only $f(x)=\frac{1}{x}+1$ works.