Let $P(x,y)$ the assertion of the given F.E., now trivially $f=0$ works so assume $f$ non-constant. Now if there existed $c \in \mathbb R$ such that $f(c)=0$ then... $P(c,x-c)$ gives $f(x)=(x-c)f(c+1)$ which means $f$ is lineal and clearly we can get $f(x)=x$ as only solution for this case (also shows that $f(c+1) \ne 0$). So now assume that $f(x) \ne 0$ for all reals $x$ then: $P(x,0)$ gives $f(x-f(x))+f(x)f(f(0))=0$. $P(x, f(x))$ gives $f(x+1)=1+f^3(x)$ which means that if $f(a)=f(b)$ then $f(a+1)=f(b+1)$. $P(x, f(x)+1)$ gives $f(x)+1=f(f(f(x)+1))$. (Involution for points of the form $f(x)+1$). Now $P(a,x)-P(b,x)$ gives that in fact $f$ is periodic with period $a-b$, but by $P(x,y+k(a-b))$ we get that $f$ is injective otherwise you get infinitely many different values for RHS. Now we have by $P(x,f(y))$ that $f(y)+f(y)f^3(x)=f(x+f(y)-f(x))+f(x)f^3(y)$, use symetry and add them to get that $f(x)+f(y)=f(x+f(y)-f(x))+f(y+f(x)-f(y))$ for all reals $x,y$. Notice we also have $f(x+1)=f^3(x)+1=f(f(f^3(x)+1)))=f^3(x+1)$ so $f(f(x+1))=x+1$ so $f$ is an involution., but we also have as well that then $f(x+1)=f(x)+1$, so now the F.E. becomes: $$y=f(x+y-f(x)) \; \forall x,y \in \mathbb R$$And trivially this means $f(x)=x$ for all $x \in \mathbb R$ which contradicts that $f$ takes no zeros as $f(0)=0$. Therefore the only solutions are $f(x)=0,x$ for all reals $x \in \mathbb R$ thus we are done

Assume for the sake of contradiction that $f$ is non-linear. Suppose $f(c)=0$. Then, $P(c,y): yf(c+1)=f(c+y)$, so $f$ is linear. If $f(a)=f(b)$, then $P(a,f(a)): f(a)f(a+1)=f(a)+f(a)f(f(f(a)))$. Combined with $P(b,f(b))$, this means that $f(a+1)=f(b+1)$. Now consider $P(a,b)$ and $P(b,a)$, which give $af(b+1)=bf(a+1)$. Hence, $a=b$, so $f$ is injective. Finally, $P(x,f(x)+1): f(x)f(x+1)=f(x)f(f(f(x)+1))$, so $f(f(x)+1)=x+1$. This means that there is some $f(c)=0$, contradiction. Hence, $f$ is linear. Checking, $f(x)=x$ or $f(x)=0$.