Let $P(x,y)$ implies $f(x + yf(x + y)) = y^2 + f(xf(y + 1))$ 1) $f(1)=1$ and $f(0)=0$ $P(x,0) \Longrightarrow f(x)=f(xf(1))$ $P(xf(x), 0) \Longrightarrow f(xf(x))=f(xf(x)f(1))$ $P(0,x) \Longrightarrow f(xf(x))=x^2+f(0)$ $P(0,xf(1)) \Longrightarrow f(xf(x))=f(xf(x)f(1))=f(xf(1)f(xf(1)))=x^2f(1)^2+f(0)$ $\Longrightarrow f(xf(x))=x^2+f(0)=x^2f(1)^2+f(0)$ $\Longrightarrow x^2=x^2f(1)^2$ So $f(1)=1$ or $f(1)=-1, \forall x \ne 0$ Case 1: $f(1)=-1$ $P(x,0) \Longrightarrow f(x)=f(-x)$ $P(1,0) \Longrightarrow f(-1)=-1$ $P(1,-2) \Longrightarrow f(3)=3$ $P(-1,2) \Longrightarrow f(3)=f(-3)=7=3$ contradiction. Case 2: $f(1)=1$ $P(0,1) \Longrightarrow f(0)=0$ So $f(1)=1$ also true when $x=0$ 2) $f(x)=1 \Longleftrightarrow x=1$ If exist $a : f(a)=1$ $P(0,a) \Longrightarrow a^2=1$ $\Longrightarrow a=1$ or $a=-1$ If $a=-1$, it implies $f(-1)=1$ $P(1;-2) \Longrightarrow 1=5$ contracdiction. So $f(x)=1 \Longleftrightarrow x=1$ 3) $f(x)=x, \forall x \in \mathbb{R}$ $P(x+1,-1) \Longrightarrow f(x+1-f(x))=1$ Since $f(x)=1 \Longleftrightarrow x=1$, it implies $x+1-f(x)=1$ And so $\boxed{f(x)=x, \forall x \in \mathbb{R}}$ Thank you Dear mathlinks.