Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$

Question

Find $f: \mathbb R \to \mathbb R$ which satisfies $f\left(xf(y)-y^2\right)=(y+1)f(x-y)$.

My attempt: \begin{align} &P(x, -1): f\bigl(xf(-1)-1\bigr)=0. \\ &\text{If } f(-1) \ne 0 \implies x \leftarrow \frac {x}{f(-1)}: f(x-1)=0 \Rightarrow f \equiv 0, \text{ which is contradiction.} \\ & \implies f(-1)=0. \\ \\ &P(0, y): f\left(-y^2\right) = (y+1)f(-y). \\ \ \\ &P(x, 0): f\bigl(xf(0)\bigr)=f(x). \\ &\text{If } f(0)=1 \implies P\left(\frac {y^2}{f(y)}, y\right): 1=f(0)=(y+1)f\left(\frac{y\bigl(y-f(y)\bigr)}{f(y)}\right) \\ &f\left(\frac{y\bigl(y-f(y)\bigr)}{f(y)}\right)=\frac {1}{y+1}. \end{align}

Answer 1

For any $y\ne -1$, choose $x$ such that $x f(y) -y^2=-1$. This is possible because, when $f$ is not constant, its only zero is $-1$. Then, $$ (y+1) f\left(\frac{y^2-1}{f(y)}-y\right) = 0\Leftrightarrow \frac{y^2-1}{f(y)}-y = -1, $$

This means that any non-constant solution satisfies $f(y)=y+1$ for any $y\ne -1$. Finally, since this candidate solution also satisfies $f(-1)=0$, it is in fact the only non-constant solution to this equation.

To conclude, this equation has two solutions: $f(x) \equiv 0$ and $f(x) = x+1$.