Find all functions $f: \mathbb{N}^{*} \rightarrow \mathbb{N}^{*}$ such that, for all $a, b \in \mathbb{N}^{*}$, we have

$$
f(y)+2 x \mid 2 f(x)+y
$$
We start by testing $x=y=1$ in the given condition: this gives that $f(1)+2$ divides $2 f(1)+1$, but also $2 f(1)+4$, so $2 f(1)+1 \mid 3$, hence $f(1)=1$.

Now we take $y=1$ to obtain, for all $x$:

$$
2 x+1 \mid 2 f(x)+1
$$

By taking $x=1$, we get for all $y$:

$$
f(y)+2 \mid y+2
$$

The first of the last two equations gives $x \leqslant f(x)$, the second gives $f(y) \geqslant y$, so $f(x)=x$ for all $x$, and it is easily verified that the identity is indeed a solution.