parmenides51 wrote: Find all functions $f: R \to R$ such that for all real numbers $x, y$ holds $f(y - xy) = f(x)y + (x - 1)^2 f(y)$ Let $P(x,y)$ be the assertion $f(y-xy)=f(x)y+(x-1)^2f(y)$ $P(3,0)$ $\implies$ $f(0)=0$ $P(1,1)$ $\implies$ $f(1)=0$ $P(x,1)$ $\implies$ $f(1-x)=f(x)$ $P(1-x,y)$ $\implies$ $f(xy)=f(x)y+x^2f(y)$ $P(1-y,x)$ $\implies$ $f(xy)=f(y)x+y^2f(x)$ Subtracting, this becomes $f(x)(y^2-y)=f(y)(x^2-x)$ And so $f(x)=c(x^2-x)$ Plugging this in original equation, we get $c=0$ And so $\boxed{f(x)=0\quad\forall x}$

I hope I am not mistaken, but I believe $f(x)=c(x^2-x)$ works for all $c \in \mathbb{R}$. $$f(y-xy)=f(x)y+(x-1)^2f(y) \Longleftrightarrow$$$$c(y-xy)(y-xy-1)=cx(x-1)y+c(x-1)^2y(y-1) \Longleftrightarrow$$$$c(-y(x-1)(y-xy-1)-x(x-1)y-(x-1)^2y(y-1))=0 \Longleftrightarrow$$$$cy(x-1)(-(y-xy-1)-x-(x-1)(y-1))=0 \Longleftrightarrow$$$$cy(x-1)(-y+xy+1-x-xy+x+y-1)=0 \Longleftrightarrow$$$$cy(x-1) \cdot 0=0$$

Let $P(x,y):=f(y - xy) = f(x)y + (x - 1)^2 f(y)$ $P(3,0)$ yields $f(0)=4f(0)\Longrightarrow f(0)=0$ $P(1,1)$ yields $f(0)=f(1)=0$ $P(-x+1,1)$ yields $f(x)=f(-x+1)$ therefore $f(y-xy)=f(xy-y+1)$ so let $Q(x,y):=f(xy-y+1)=f(x)y+(x-1)^2f(y)$ $Q(1-1/x,x)$ for $x\neq0$ yields $f(-1/x+1)+\frac{1}{x^2}f(x)=0$ Now using the fact that $f(x)=f(-x+1)$ we obtain $f(1/x)+\frac{1}{x^2}f(x)=0\Longleftrightarrow f(x)=-x^2f(1/x), \forall x\neq0$ Also now now note that letting $x\to-1$ in our last equation yields $f(-1)=0$ $P(x+1,1/x)$ for $x\neq0$ yields $f(1/x-1/x(x+1))=f(x+1)1/x+x^2f(1/x)=f(x+1)1/x-f(x)$ Furthermore using $f(x)=f(-x+1)$ in the form $f(-x)=f(x+1)$ we can transform the equation into $f(-1)=f(-x)1/x-f(x)$ Now using that $f(-1)=0$ we obtain $f(-x)1/x-f(x)=0\Longrightarrow xf(x)=f(-x), \forall x\neq0$ let this assertion be $T(x)$ $T(-x)$ yields $-xf(-x)=f(x)$ furthermore plugging this into $T(x)$ yields $-x^2f(-x)=f(-x)\Longrightarrow f(-x)(x^2+1)=0$ as $x^2+1\ge1$ we must have $f(-x)=0, \forall x\neq0$ which is equivalent to $f\equiv 0, \forall x\neq0$ however as $f(0)=0$ we obtain that $f\equiv 0, \forall x\in\mathbb{R}$ In conclusion $\boxed{f\equiv0, \forall x\in\mathbb{R}}$ $\blacksquare$.