Find all Function $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+yf(x))=f(x)+xf(y) \; \forall \; x,y \in \mathbb R$

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Find all Function $f:\mathbb R \rightarrow \mathbb R$ such that $f(x+yf(x))=f(x)+xf(y) \; \forall \; x,y \in \mathbb R$

My Approach:

Let $P(x,y):f(x+yf(x))=f(x)+xf(y)$

Set $x=1,y=0$

$P(1,0): f(1)=f(1)+f(0)$

$\implies$

$f(0)=0$.

After this step i tried many combinations of $x$ and $y$. But I couldn't Find solution.

$f(x)=0$ seems a obvious solution. I also tried $f(x)=x$ because many functional equation has solution $f(x)=x$ and that solution worked but I don't know how to prove this (i.e., $f(x)=x$) solution and any other solution which given equation can have.