$P(1,y)$ gives us $f(f(2y))=y(1+f(1))$. $P(f(2x),y)$ gives us $f(f(2x)f(2y))=\frac{y}{f(2x)}+yf(f(2x))=\frac{y}{f(2x)}+yx(1+f(1))$. By symmetry, we get $\frac{y}{f(2x)}+yx(1+f(1))=\frac{x}{f(2y)}+xy(1+f(1))\Rightarrow 2xf(2x)=2yf(2y)\Rightarrow xf(x)=yf(y)$ for all $x,y\in\mathbb{R^+}$. Hence, $f(x)=\frac cx$. Plugging in the original equation yields $c=1$ so $f(x)\equiv \frac 1x$ is the only solution.

Let me provide a slicker solution. As Baris did, $P(1,y)$ gives $f(f(2y)) = y(1+f(1))$. In particular, $f$ is surjective on $\mathbb{R}^+$. Now, let $y_1$ be such that $f(2y_1)=1$. Then, $P(x,y_1)$ gives $xf(x) = y_1 + y_1xf(x)$, that is $xf(x)(1-y_1)=y_1$. Now that $y_1\ne 1$ clearly, we have $f(x)=c/x$ for some constant $c>0$. Then get $f(x)=1/x$ immediately.

\[xf(xf(2y))=y(xf(x)+1)\]If $f(a)=f(b),$ then plugging $y=\frac{a}{2},\frac{b}{2}$ gives contradiction thus, $f$ is injective. $P(\frac{1}{f(1)},\frac{1}{2})\implies 1=\frac{1}{2}(\frac{f(\frac{1}{f(1)})}{f(1)}+1)\iff f(1)=f(\frac{1}{f(1)})\overset{\text{by injectivity}}{\implies} f(1)=1$ $P(x,\frac{1}{2})\implies xf(x)=\frac{1}{2}(xf(x)+1)\iff xf(x)=1\iff \boxed{f(x)=\frac{1}{x}, \ \forall x\in R^+}$