If $f \not \equiv 0$: $P(\frac{x-f(f(y))}{f(y)},y): f(x)=x+f(y)-f(f(y))\Rightarrow f(f(y))=f(y)+f(y)-f(f(y))\Rightarrow f(f(y))=f(y)\Rightarrow f(x)\equiv x$.

parmenides51 wrote: Find all functions $f : R \to R$ such that $$f(xf(y) + f(f(y))) = (x + 1)f(y)$$for all real numbers $x$ and $y$. Let $P(x,y)$ be the assertion $f(xf(y)+f(f(y)))=(x+1)f(y)$ $\boxed{\text{S1 : }f(x)=0\quad\forall x}$ is a solution. If $f\not\equiv 0$, then $(x+1)f(y)$ can take any value we want and so $f(x)$ is surjective and $P(x,y)$ may be written : New assertion $Q(x,y)$ : $f(xy+f(y))=(x+1)y$ $Q(0,x)$ $\implies$ $f(f(x))=x$ and $f(x)$ is also injective. $Q(-1,x)$ $\implies$ $f(f(x)-x)=0$ and so, since injective, $f(x)=x+a$ for some constant $a=f^{-1}(0)$ Plugging this back in original equation, we get $a=0$ and so $\boxed{\text{S2 : }f(x)=x\quad\forall x}$,which indeed fits.

We claim that the answer is \[f\equiv 0 \text{ and } f(x)=x \text{ for all }x \in \mathbb{R}\]It is easy to see that these functions satisfy the given equation. Now we shall show that they are the only ones. In what follows assume that $f\not \equiv 0$. First note that, $P(0,y)$ gives \[f(f(f(y)))=f(y)\]for all $y \in \mathbb{R}$. Then, \[2f(f(y))=f(f(f(y))+f(f(f(y))))=f(f(f(y))+f(y))=2f(y)\]Thus, $f(f(y))=f(y)$ for all $y \in \mathbb{R}$. Now, note that if $f$ is not constant zero, there exists $\alpha$ such that $f(\alpha)=c\neq 0$. Thus, for all $t\in \mathbb{R}$, we have that \[f\left( \frac{t-c}{c}f(\alpha) +f(f(\alpha)) \right)=\left( \frac{t-c}{c}+1 \right)f(\alpha)=t\]Thus, for all $t\in \mathbb{R}$, there exists some $t' \in \mathbb{R}$ such that $f(t')=t$ which immediately implies that $f$ is surjective. Thus, using the previously obtained fact that $f(f(y))=f(y)$ for all $y \in \mathbb{R}$, we must have that, \[f(x)=x\]for all $x\in \mathbb{R}$ as desired.