Let's determine all functions $f(f(x))=x+f(x)$ for all real numbers $x$.
First, we prove some properties of the function $f$ that follow from the functional equation and the continuity of $f$.

i) $f$ is bijective. If $f\left(x_{1}\right)=f\left(x_{2}\right)$, then of course $f\left(f\left(x_{1}\right)\right)=f\left(f\left(x_{2}\right)\right)$ also holds, so according to the functional equation, $x_{1}=f\left(f\left(x_{1}\right)\right)-f\left(x_{1}\right)=f\left(f\left(x_{2}\right)\right)-f\left(x_{2}\right)=x_{2}$, hence $f$ is indeed bijective.

ii) $f$ is either strictly increasing or strictly decreasing, since if a continuous function is bijective, then it is strictly increasing or decreasing.

iii) $f(0)=0$, indeed if $x=0$, then according to the functional equation $f(f(0))=0+f(0)$, the bijective function takes equal values at 0 and $f(0)$. This is only possible if $f(0)=0$.

iv) If $f$ is strictly increasing, then its range is the set of real numbers. Indeed, if $x>0$, then $f(x)>f(0)=0$, so $f(f(x))=x+f(x)>x$, hence the function $f$ also takes values greater than any positive number - and similarly, it takes values less than any negative number. Thus, since it is continuous, its range is the set of real numbers. Its inverse, $f^{-1}$, is therefore defined everywhere and is also strictly increasing.

Now let's turn to the solution.

First, consider the case when the function $f$ is strictly decreasing. Let $x \neq 0$ be arbitrary, and consider the sequence $x_{1}=f(x), x_{2}=f(f(x))=f\left(x_{1}\right), \ldots, x_{n}=f\left(x_{n-1}\right), \ldots$. If $x=x_{0}$, then according to the functional equation, $x_{k}+f\left(x_{k}\right)=f\left(f\left(x_{k}\right)\right)$, i.e., $x_{k}+x_{k+1}=x_{k+2}$ if $k \geq 0$. The sequence $x_{k}$ is thus a Fibonacci-type sequence. It is known - see, for example, the article by Kós Géza and Énekes Béla in the 9th issue of KöMaL in 2000 - that Fibonacci-type sequences can be written using two special such sequences: there exist numbers $c_{1}, c_{2}$ such that

$$
x_{n}=c_{1} \cdot \tau^{n}+x_{2}(-\tau)^{-n}, \quad \text { where } \tau=\frac{1+\sqrt{5}}{2}
$$

Since $f$ is strictly decreasing and $f(0)=0$, $x$ and $f(x)$ have opposite signs, and the consecutive terms of the above $x_{n}$ sequence also do. In the above (1) representation, the first term, $c_{1} \tau^{n}$, has a constant sign, and its absolute value can be arbitrarily large as $n$ increases. The second term, $c_{2}(-\tau)^{-n}$, has alternating signs, but this sequence tends to 0 as $n$ tends to infinity. The sequence $x_{n}$ can only be alternating in sign like the second term if the constant sign "dominant term", $c_{1} \cdot \tau^{n}=0$, i.e., $c_{1}=0$.

Then $x_{n}=c_{2} \cdot(-\tau)^{-n} ; x_{0}=c_{2}(-\tau)^{0}=c_{2}=x$, so $f(x)=x_{1}=x \cdot(-\tau)^{-1}$, i.e., if $f$ is decreasing, then only

$$
f(x)=-\frac{x}{\tau}
$$

is possible, where $\tau=\frac{1+\sqrt{5}}{2}$. It can be easily verified - using, for example, the relation $\tau^{2}=\tau+1$ - that this continuous function satisfies the given functional equation.

Now let $f$ be strictly increasing. According to iv), $f^{-1}$ is defined everywhere, so starting from $x \neq 0$, the sequence $x_{-1}=f^{-1}(x), x_{-2}=f^{-1}\left(x_{1}\right), \ldots, x_{-n}=f^{-1}\left(x_{-n+1}\right), \ldots$ can be constructed; here we build the sequence by stepping backwards from the values of $f$.

$$
\ldots x_{-3} \xrightarrow{f} x_{-2} \xrightarrow{f} x_{-1} \xrightarrow{f} x_{0} \xrightarrow{f} x_{1} \xrightarrow{f} x_{2} \xrightarrow{f} \ldots
$$

By extending the previous $x_{n}$ sequence with the negative index terms, we still get a Fibonacci-type sequence, and the (1) relation remains valid for the $n$-th term of the extended sequence as well.

If now $n \rightarrow-\infty$, then of course the alternating sign $c_{2}(-\tau)^{-n}$ will be the "dominant", $c_{1} \cdot \tau^{n}$ will tend to 0. The sequence $x_{n}$ $(n \in \mathbf{Z})$ is now of constant sign, since if $x>0$ then $f(x)>0$, and if $x<0$ then $f(x)<0$. This is only possible if the unbounded, alternating sign $c_{2}(-\tau)^{-n}$ part is 0, $x_{n}=c_{1} \tau^{n}$; from which, similarly to the previous case, $f(x)=x_{1}=x \cdot \tau^{1}$, i.e.,

$$
f(x)=\tau x
$$

follows. From the relation $\tau^{2}=\tau+1$, it immediately follows that this function is also a solution.

Thus, the functional equation has two continuous solutions; both are linear functions,

$$
f_{1}(x)=-\frac{x}{\tau} \quad \text { and } \quad f_{2}(x)=\tau x
$$

where $\tau=\frac{1+\sqrt{5}}{2}$.