$f: \Bbb{R} \to \Bbb{R}, f(y)f(x+f(y))=f(xy)+f(y)^2.$
\begin{align} P(0, 0): \; & f(0)f(f(0))=f(0)+f(0)^2. \ \\ \text{if } \; & f(0) \neq 0: \\ & f(f(0))=f(0)+1. \\ \ \\ P(-1, f(0)): \; & (f(0)+1)^2=f(-f(0))+(f(0)+1)^2. \\ \therefore \; & f(-f(0))=0. \\ P(0, -f(0)): \; & 0=f(0), \text{Contradiction.} \\ \ \\ \therefore \; & f(0)=0. \\ \ \\ P(0, y): \; & f(y)f(f(y))=f(y)^2. \\ \text{if } \; & f(y) \neq 0: \\ & f(f(y))=f(y). \end{align}
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As you noted in comments, the solution is either $f(x)\equiv 0$, or $x=0$ is the unique zero of $f$. For the second case now notice that $P(y-f(y),y)$ gives after some cancellation $$ 0=f(y(y-f(y))). $$ So by the uniqueness of zero we have $y(y-f(y))=0$, hence for $y\neq 0$ we get $f(y)=y$. Together with $f(0)=0$ this gives the identity function as the only remaining solution.
This is basically the cancellation method: we tried to find values of $x,y$ such that some terms on both sides become equal and cancel. In this case we would try $x+f(y)=xy$, or $x+f(y)=y$ where the second one leads to the solution above.