Determine all functions $f: \mathbb{R}_{+} \rightarrow \mathbb{R}_{+}$ such that for all $x \geqslant 0$, we have $f(f(x)) +$ $f(x)=6 x$.
Let $x \geqslant 0$. We define the sequence $\left(x_{n}\right)$ by $x_{0}=x$ and $x_{n+1}=f\left(x_{n}\right)$. According to the functional equation, this sequence satisfies the recurrence relation $x_{n+2}=-x_{n+1}+6 x_{n}$. The characteristic polynomial, $t^{2}+t-6$, has two distinct roots -3 and 2, so there exist real numbers $\lambda$ and $\mu$ such that for all $n, x_{n}=\lambda \cdot(-3)^{n}+\mu \cdot 2^{n}$. Suppose $\lambda$ is non-zero; then for large $n$, $x_{n}$ has the same sign as $\lambda \cdot(-3)^{n}$. If $\lambda>0$, then $x_{n}$ is negative for odd $n$, and if $\lambda<0$, then $x_{n}$ is negative for even $n$. In both cases, we reach a contradiction, so $\lambda=0$. Using the fact that $x_{0}=x$, we find that $\mu=x$. Therefore, $f(x)=x_{1}=2 x$.

Conversely, $f(x)=2 x$ is indeed a solution to the functional equation.