Set $x=y=0$ yields $f(0)=f^2(0)\implies f(0) = 0, 1 $. 1) If $f(0)=0$, setting $y=0$ leads to $f(-x)=2x \implies f(x)={-2x}$. 2) If $f(0)=1$, setting $y=0$ leads to \begin{equation} f(-x)=f(x)+2x \end{equation} Replacing $x$ with $(yf(x)-x)$ in the \myJustRef{eq_jpn_mo_1} leads to \begin{align*} f(x-yf(x)) &= f(y(f(x)-x) + 2(yf(x)-x) \\ &= (f(x)f(y) + 2x) + 2(yf(x)-x) &\because \\ &= f(x)f(y) + 2yf(x)\\ &= f(x)(f(y)+2y) \\ &= f(x)f(-y) &\because\\ \\ \therefore\qquad& f(x-yf(x))=f(x)f(-y) \end{align*} Replacing $y$ with $-y$ and note () gives \begin{align*} &f(x+yf(x))&=f(x)f(y)\implies& f(x+yf(x))&= f(yf(x)-x)-2x \end{align*} Setting $y=-\frac{x}{f(x)}$ in the last relation yields $$f(0)=f(-2x)-2x\implies 1=f(-2x)-2x \implies f(-2x)=1-(-2x)$$ Finally, replacing $-2x$ with $x$ leads to another solution $$f(x)={1-x}$$ Therefore, there exist totally two solutions $f(x)=\boxed{-2x}$ or $\boxed{1-x}$.