Let $P(x,y): f(xy+f(x))=xf(y)+f(x)$ $P(x,0): f(f(x))=xf(0)+f(x)$.If $f(x_1)=f(x_2)$ then $x_1f(0)=x_2f(0)$ so $f(0)=0$ or $f$ is $1-1$. Case 1:$f(0)=0$ $f(f(x))=f(x)\,\,(*)$ $P(x,\dfrac{-f(x)}{x}): 0=xf(-\dfrac{f(x)}{x})+f(x),x\in \mathbb{R^*}$ Let $S=\left \{x\in \mathbb{R}:f(x)=0 \right \}$.If $S=R$ then $f(x)=0,\forall x\in \mathbb{R}$ which is a accepted solution. If there exist $x_1$ such that $f(x_1)\neq 0$ then $P(f(x_1),\dfrac{-f(f(x_1))}{f(x_1)}):0=f(x_1)f(\dfrac{-f(f(x_1))}{f(x_1)})+f(f(x_1))\overset{(*)}{\Leftrightarrow} 0=f(x_1)(f(\dfrac{-f(x_1)} {f(x_1)})+1)\Leftrightarrow f(-1)=-1$ $0\in S$,suppose that there exist $x_2\in S,x_2\neq 0$. $P(x_2,1): f(x_2)=x_2f(1)\Leftrightarrow f(1)=0$ $P(-1,x):f(-x-1)=-f(x)-1$. $P(-1,1):f(-2)=-1$ $P(-2,-1):f(2+f(-2))=-2f(-1)+f(-2)\Leftrightarrow f(1)=2+f(-2)=1\Leftrightarrow 0=1$,contradiction So $S=\left \{0\right \}$ $P(x,\dfrac{x-f(x)}{x}):f(x)=xf(\dfrac{x-f(x)}{x})+f(x)\Leftrightarrow \dfrac{x-f(x)}{x}=0\Leftrightarrow f(x)=x,\forall x\neq 0$.Also $f(0)=0$ so $f(x)=x \forall ,x\in \mathbb{R}$,which is a accepted solution. Case 2: $f(0)\neq 0\Rightarrow f:1-1$ $P(0,0):f(f(0))=f(0)\Leftrightarrow f(0)=0$,contradiction So the solutions are $f(x)\equiv 0\forall x \in \mathbb{R}$ ,$f(x)=x\forall x\in \mathbb{R}$