(F-M) Let $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $x+f(x)=f(f(x))$ determine the solutions of $f(f(x))=0$.
We have the injectivity of $f$ and since $f(x)+f^{2}(x)=f^{3}(x)=f(x+f(x))$ we have $f(f(x))=0$ if and only if $f(x)=f(x+f(x))$ if and only if $x=x+f(x)$ if and only if $f(x)=0$ and thus $x+f(x)=f(f(x))=x=0$ so the only potential solution is 0. Furthermore, since $f(0)=f(f(0))$ we have $f(0)=0$ and thus $f(f(0))=0$.