$\displaystyle solve f(x)+2x=f(f(x))$
Solve using some knowledge of "involuntary functions." For example, if for nonzero $\displaystyle x,$ $\displaystyle f(x)=1/x,$ then $\displaystyle f(f(x))=x$ again. Similarly, if $\displaystyle f(x)=-x,$ then $\displaystyle f(f(x))=x$ again. This last one is what we want. So
$\displaystyle f(x)=-x \implies f(x)+2x=x=f(f(x)),$
Q.E.D.
Just some stuff I've found out about the equation...
Observation 1 : We write: $\displaystyle f^k(x)=f(f(...f(x)))$ ( where f appears k times )
Your equation then turns into: $\displaystyle f^1(x)+2f^{0}(x)=f^2(x)$ very suggestive, isn't it?
This clearly generalises to: $\displaystyle f^{n+1}(x)+2f^{n}(x)=f^{n+2}(x)$ now fix $\displaystyle x$ and set: $\displaystyle a_n=f^n(x)$ solving the recurrence relation: $\displaystyle f^k(x)=A(x)\cdot{(-1)^k}+B(x)\cdot{2^k}$
Setting: $\displaystyle k=0$ we have the relation: $\displaystyle x=A(x)+B(x)$ ( in particular your solutions correspond to the cases $\displaystyle A(x)=0$ and the other $\displaystyle B(x)=0$ )
Now, this implies: $\displaystyle f(x)=-A(x)+2B(x)=-x+3B(x)$ and going to the original equation we get the following restriction $\displaystyle 2B(x)=B(3B(x)-x)=B(f(x))$
So it seems it all comes down to solving that functional equation.
Observation 2: $\displaystyle f$ is injective. To see this, suppose $\displaystyle f(x)=f(y)$ then $\displaystyle f(f(x))=f(f(y))$ thus: $\displaystyle f(x)+2x=f(y)+2y$ but then $\displaystyle 2x=2y$ and so $\displaystyle x=y$
Observation 3: It follows from 2 that since $\displaystyle f(f(0))=f(0)$ we must have $\displaystyle f(0)=0$