Find the functions $f: \mathbb{R}_{+}^{*} \rightarrow \mathbb{R}_{+}^{*}$ satisfying, for all $x, y>0$,

$$
f(2 x f(3 y))+f\left(27 y^{3} f(2 x)\right)=6 x y+54 x y^{3}
$$

Solved by Gä̈tan Dautzenberg and Georges Tézé
Solved by Gaëtan Dautzenberg and Georges Tézé. We see that the identity is a solution.

Let $f$ be a solution: $\forall x, y>0, f(x f(y))+f\left(y^{3} f(x)\right)=x y+x y^{3}$.

We make the substitutions

- $x=y=1: f(f(1))=1$
- $x=f(1), y=1: f\left(f(1)^{2}\right)=f(1)$
- $x=f(1)^{2}, y=1: f\left(f(1)^{3}\right)=2 f(1)^{2}-1$

- $x=y=f(1): f(1)^{2}=f(1)^{4}$, so $f(1)=1$.

Thus, substituting $y=1$, we have $\forall x>1, f(x)+f(f(x))=2 x$.

We then show by induction that if $x>0$ is fixed, we have

$$
\forall n \in \mathbb{N}, f^{n}(x)=\frac{2+(-2)^{n}}{3} x+\frac{1-(-2)^{n}}{3} f(x)
$$

where $f^{n}$ is the $n$-th iterate of $f$.

Thus, since $f>0$,

$$
0<f^{2 n}(x)=\frac{2^{2 n}+2}{3} x-\frac{2^{2 n}-1}{3} f(x) \text { and } 0<f^{2 n+1}(x)=\frac{2^{2 n+1}+1}{3} f(x)-\frac{2^{2 n+1}-2}{3} x
$$

Therefore, for all $n \geqslant 1$, we have

$$
\frac{2^{2 n+1}-2}{2^{2 n+1}+1}<\frac{f(x)}{x}<\frac{2^{2 n}+2}{2^{2 n}-1}
$$

By taking the limit as $n$ approaches infinity, we find $1 \leqslant \frac{f(x)}{x} \leqslant 1$ so $f(x)=x$.