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Setting $x=y$ gives $0=f(0)(f(2x)-1)$, $\forall x\in \mathbb{R}$.

First case: $f(0)\ne 0$ We have $f(2x)-1=0$, $\forall x\in \mathbb{R}$, so $f(x)=1$, $\forall x\in \mathbb{R}$. We check that this function is a solution.

Second case: $f(0)=0$ Setting $y=0$ gives $f(f(x))-f(x)=f(x)(f(x)-1)$, i.e. $$f(f(x))=(f(x){)}^2,\forall x\in \mathbb{R}.(\star )$$ Interchanging $x$ and $y$ gives $f(x+f(y))-f(y+f(x))=f(y-x)(f(x+y)-1)$, which added to the starting equation gives $$0=(f(x+y)-1)(f(x-y)+f(y-x)),\forall x,y\in \mathbb{R}.$$ Setting $y=-x$ in thus obtained equation gives $0=(f(0)-1)(f(2x)+f(-2x))$, so $f(0)=0$ implies $f(2x)=-f(-2x)$, $\forall x\in \mathbb{R}$ and hence the function $f$ is odd.

Now we conclude $$f(f(x))\stackrel{\ast }{=}(f(x){)}^2=(-f(-x){)}^2=(f(-x){)}^2\stackrel{\ast }{=}f(f(-x))=-f(-f(-x))=-f(f(x)),(\star \star )$$ for every $x\in \mathbb{R}$

Finally, $(\star \star )$ and $(\star )$ give $2f(f(x))=0$, so $2(f(x){)}^2=0$. We check directly that $f(x)=0$, $\forall x\in \mathbb{R}$ is also a solution.