Denote $P(x,y)$ by the assertion of $f(xy+f(x))=yf(x)+x$. $P\left(x,\dfrac{y}{f(x)}\right) \Rightarrow f\left(\dfrac{xy}{f(x)}+f(x)\right)=y+x \forall x,y>0$ This implies $f$ is surjective on $\mathbb{R}^+$. $\forall y>f(x), P\left(x,\dfrac{y-f(x)}{x}\right) \Rightarrow f(y)=f(x) \cdot \dfrac{y-f(x)}{x}+x$ Since $f$ is surjective on $\mathbb{R}^+$, the equality is true for all $y>0$ and $x \in \mathbb{R}^+$ that $f(x)<y$. Therefore, we have $f$ is injective, or $f$ is bijective on $\mathbb{R}^+$ $P(x,1) \Rightarrow f(x+f(x))=x+f(x) \forall x>0$ $(1)$ $P(x+f(x),y) \Rightarrow f((y+1)(x+f(x))=(y+1)(x+f(x)) \forall x,y>0$ $(2)$ Let $A=\lbrace x \in \mathbb{R}^+ \mid f(x)=x \rbrace$ From $(1)$ we have $x+f(x) \in A \forall x > 0 \Rightarrow A \neq \varnothing$ From $(2)$ we have $x(y+1) \in A \forall x \in A, y>0$ $\Rightarrow x \in A \forall x \geq x_0$ with $x_0$ is an element of $A$. $(3)$ Let $0<x<x_0$ and $y$ big enough that $xy+f(x)> x_0$ and $y>1$. $P(x,y) \Rightarrow xy+f(x)=yf(x)+x$ $\Rightarrow (f(x)-x)(y-1)=0 \forall x<x_0$ Since $y>1$, we have $f(x)=x \forall x<x_0$. Combine with $(3)$ we have $f(x)=x \forall x>0$. Hence, $f(x)=x$ is the only solution.

Let the assertion $P(x,y)=f(xy+f(x))=yf(x)+x$. $P(x,1)$ yields $f(x+f(x))=x+f(x)$, this implies that $\exists f(n)=n$ for some $n \in \mathbb{R}^+$. Furthermore $P(n,\frac{1}{f(n)})$ yields: $f(n+1)=n+1$, thus $\exists f(k)=k>n$ Now let's choose an arbitrary $x$ and $y$ such that $y>1$. Furthermore we have: $xy+f(x)=f(xy+f(x))=yf(x)+x \Longleftrightarrow xy-x-yf(x)+f(x)=0 \implies x(y-1)-f(x)(y-1)=(y-1)(x-f(x))=0$ Thus $f(x)=x$. However since our choice was arbitrary we can easily extend the solution to all positive reals which implies that $\boxed{f(x)=x, \forall x\in \mathbb{R}^+}$ QED. And we are done!

Let $P(x,y)$ be the assertion that $f(xy+f(x))=yf(x)+x.$ Let $f(1) = a$ $P(1, x-a)$ gives us that $(*)f(x) = ax - a^2 + 1$ for all $x > a$ $P(1,1)$ gives us that $f(a+1) = a+1$ $P(a+1,\tfrac{x-a-1}{a+1})$ gives us that $(**)f(x) = x$ for all $x > a + 1$ Comparing $(*)$ with $(**)$, we see that $a=1$, and $f(x) = x$ for all $x>1$ Finally, taking $P(x,\tfrac{1}{x})$ gives us that $\boxed{f(x) = x}$ for all $x>0$