$P(1,1): f(1)=f(f(1))$ $P(f(1),1): f(1)^2 (2f(1))=(f(1)+1)f(1) \implies 2f(1)^3-f(1)^2-f(1)=0 \implies f(1)= 0$, $-\frac{1}{2}$ or $1$. Since $f$ is over positive reals, $f(1)=1$ $P(1,x): f(x)+1=(x+1)f(x) \implies f(x)=\frac{1}{x}$