Let $P(x,y)$ the given condition $P(0,0) \Rightarrow f(0)=0$. $P(x,-x): 0=f(x^2)+xf(-x)$ and by $x\to -x$ we get that $f(-x)=-f(x),\forall x\in \mathbb{R}$ and $f(x^2)=xf(x)$ Now let $Q(x,y)=P(x-y,y): f(x+y(f(x)-1))=xf(y)+f(x-y)$. Suppose that there exist $r\neq 0$ such that $f(r)=0$. Then $Q(r,x): f(r-x)=rf(x)+f(r-x)\Leftrightarrow \boxed{f\equiv 0}$ which fits. Now let $f(u)=0\Leftrightarrow u=0$. $Q(1,x): f(1+x(f(1)-1))=f(x)+f(1-x)$. The equation $1+x(f(1)-1)=x $ has a real solution if and only $f(1)\neq 2$. So, if $f(1)\neq 2$ then take $t$ such that $1+t(f(1)-1)=t$ and now $Q(1,t): f(1-t)=0$ so $t=1$ and $1+f(1)-1=1\Leftrightarrow f(1)=1$ $Q(1,x): 1=f(x)+f(1-x)\Leftrightarrow 1+f(x-1)=f(x)\Leftrightarrow f(x+1)=f(x)+1$ For $x\neq 0$ consider $Q(x,\dfrac{1}{f(x)}): f(x-1/f(x)+1)=f(1/f(x))+f(x-1/f(x))\Leftrightarrow f(1/f(x))=1/x$ so $f$ is injective. But $P(0,x): f(xf(x))=f(x^2)$ so $\boxed{f(x)=x}$ It remains the case where $f(1)=2$. Then $Q(1,x): f(x+1)=f(x)+f(1-x)\Leftrightarrow f(x+1)+f(x-1)=f(x)$. $x\to x+1: f(x+2)+f(x)=f(x+1)$ and adding this to the previous we get $f(x-1)+f(x+2)=0\Leftrightarrow f(x)+f(x+3)=0$. Now $x\to x+3: f(x+3)+f(x+6)=0$ so $f(x)=f(x+6)$. $P(x+6,y)-P(x,y):xf(y)=(x+6)f(y)\Leftrightarrow f\equiv 0$ which is impossible since $f(1)=2\neq 0$ etc

My solution is a tiny bit different than #2 (hi Prod55! ). Obtain that $f(0)=0$, $f$ is odd and $f(x^2)=xf(x)$ as above. If $f$ is constant, then $f$ is zero everywhere. Assume now that $f$ is non-constant. Claim 1: $f(x)=0 \Leftrightarrow x=0$. Proof: If $f(u)=0$ for some $u$, then take $y=u-x$ to obtain that $uf(u-x)=0$, and so $u=0$, as desired $\blacksquare$ Claim 2: $f(1)=1$. Proof: Take $x=y=-1/2$. Then, $f(-\dfrac{f(-1)}{2}-\dfrac{1}{2})=0,$ and so by Claim 1 $f(1)=-1$, and since $f$ is odd we conclude that $f(1)=1$ $\blacksquare$ Claim 3: $f(x+1)=f(x)+1$ for all $x$. Proof: Take $y=1-x$. Then, $1=f(1-x)+f(x),$ and since $f$ is odd $f(x)=f(x-1)+1,$ and so $f(x+1)=f(x)+1$ $\blacksquare$ Claim 4: $f$ is injective. Proof: Take $x=t-\dfrac{1}{f(t)}$ and $y=\dfrac{1}{f(t)}$ for some $t \neq 0$.. Then, $f(t-\dfrac{1}{f(t)}+1)=tf(\dfrac{1}{f(t)})+f(t-\dfrac{1}{f(t)}),$ and so by Claim 3 we conclude that $f(\dfrac{1}{f(t)})=\dfrac{1}{t},$ and combining this with Claim 1 we deduce that $f$ is injective $\blacksquare$ To the problem, since $f$ is injective and $x=0$ in the original equation gives $f(yf(y))=f(y^2),$ we conclude that $f$ is the identity function, which evidently works. To sum up, all solutions are the zero and the identity functions.

Let $P(x,y)$ be the assertion of: $f(x+yf(x+y))=f(y^2)+xf(y)+xf(y)+f(x)$ $P(0,0)\implies f(0)=0$ $P(0,x)\implies f(xf(x))=f(x^2)$ (1) $P(-x,x)\implies f(x^2)=xf(x)$ (2) From (1) and (2) we get: $f(xf(x))=xf(x)$ Case 1:When the function is constant: Claim:$f(x)=0$

Proof

Case 2:When the function is non-constant: Claim:$f(x)=x$

Proof

And we are done!

The answer is $f\equiv 0$ and $f(x)\equiv x$ only. They clearly work. Now I show they are the only functions that work. $P(x,0)$ gives $f(0)=0$ $P(0,y)$ gives $f(yf(y)) = f(y^2)$. This allows us to write$$f(x+yf(x+y)) =(x+y)f(y) + f(x)$$$P(-y,y)$ gives $f(y^2)=yf(y)=f(yf(y))$. This implies that $f$ is odd since $yf(y)=-yf(-y)$ Since the $x+y$ is wrapped, we re-substitute $x=s-a, y=a$ to get $$f(s-a+af(s)) = sf(a)+f(s-a)$$Claim: $f(1)=1$ or $f\equiv 0$ Proof: we let $Q(s,a)$ denote the above assertion. $Q(1,a)$ gives $$f(1-a)+f(a)=f(1+(f(1)-1)a) = f(1+(f(1)-1)(1-a))$$This implies that if $a+b=f(1)+1$ then $f(a)=f(b)$. In particular, $f(f(1)+1)=0$. If f(t)=0 for some $t\ne 0$ then $Q(t,a)$ implies $f\equiv 0$. therefore $f(1)=-1$ and now plugging $a=\frac 12$ gives $f(1/2)=0$ Now $Q(1,a)$ gives $f(a)+f(1-a)=1$. This implies $f(a+1)=f(a)+1$ Now $Q(s+1,a)$ gives $f(s+af(s))=(s+1)f(a)+f(s-a)$ Compare with $Q(-s,a)$ and $f$ odd suggests $2f(a)+f(s-a)=f(s+a)$. Plugging $s=a$ suggests $2f(a)=f(2a)$ so $f$ is additive. Now I need to deal with $f(af(s))=sf(a)$ so $f$ is involution; $f(yf(y))=f(y^2)$ imply $f$ is identity.