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}