Find all functions $f, g: \mathbb{R}_{+}^{*} \rightarrow \mathbb{R}_{+}^{*}$ such that

$$
\forall x>0, \quad\left\{\begin{array}{l}
g(f(x))=\frac{x}{x f(x)-2} \\
f(g(x))=\frac{x}{x g(x)-2}
\end{array} .\right.
$$
Notice that since the left side of the equation is necessarily strictly positive, we always have $x f(x)-2>0$ and $x g(x)-2>0$ for all $x>0$, or equivalently $x f(x)>2$ and $x g(x)>2$ for all $x$.

Suppose now that we have shown that for all $x, x f(x)>a$ and $x g(x)>a$, where $a>1$ is a strictly positive constant. By reintroducing into the equations, we find that for all $x>0$,

$$
\begin{gathered}
\frac{x}{x f(x)-2}=g(f(x))>\frac{a}{f(x)} \\
x f(x)>a x f(x)-2 a
\end{gathered}
$$

and thus

$$
x f(x)1$. The same reasoning gives that for all $x>0, x g(x)0$,

$$
\begin{gathered}
\frac{x}{x f(x)-2}=g(f(x))\frac{4 a}{a+1}
\end{gathered}
$$

and the same inequality for $g$. Let us set $a_{0}=2$ and for all $n, a_{n+1}=\frac{4 a_{n}}{a_{n}+1}$. It is easily verified that for all $n, a_{n}>1$, so we know that for all $n \geq 0$ and $x>0$,

$$
a_{n}\frac{4 a_{n}}{3+1}=a_{n}$, hence the sequence $a_{n}$ is increasing. Thus, since it is bounded by 3, it converges, to 3 which is the unique fixed point of $x \mapsto \frac{4 x}{x+1}$ greater than $a_{0}=2$.

In the limit, we therefore obtain that for all $x, x f(x) \geq 3$. But furthermore, $\frac{2 a_{n}}{a_{n}-1}$ also converges, to $\frac{2 \cdot 3}{3-1}=3$, and thus in the limit, $x f(x) \leq 3$ for all $x$. Since the same applies to $g$, we have that for all $x>0$,

$$
f(x)=g(x)=\frac{3}{x}
$$

Conversely, it is easily verified that this pair of functions is indeed a solution.