Find all functions \( f: \mathbb{R}^{+} \longrightarrow \mathbb{R}^{+} \) such that

\[
f(x f(x+y))=y f(x)+1
\]

holds for all \( x, y \in \mathbb{R}^{+} \).

## Proposed by North Macedonia
1. We will show that $f(x)=\frac{1}{x}$ for every $x \in \mathbb{R}^{+}$. It is easy to check that this function satisfies the equation.

We write $P(x, y)$ for the assertion that $f(x f(x+y))=y f(x)+1$.
We first show that $f$ is injective. So assume $f\left(x_{1}\right)=f\left(x_{2}\right)$ and take any $x1$ there is an $x$ such that $f(x)=z$. Indeed $P\left(x, \frac{z-1}{f(x)}\right)$ gives that

$$
f\left(x f\left(x+\frac{z-1}{f(x)}\right)\right)=z .
$$

Now given $z>1$, take $x$ such that $f(x)=z$. Then $P\left(x, \frac{z-1}{z}\right)$ gives

$$
f\left(x f\left(x+\frac{z-1}{z}\right)\right)=\frac{z-1}{z} f(x)+1=z=f(x) .
$$

Since $f$ is injective, we deduce that $f\left(x+\frac{z-1}{z}\right)=1$.
So there is a $k \in \mathbb{R}^{+}$ such that $f(k)=1$. Since $f$ is injective this $k$ is unique. Therefore $x=k+\frac{1}{z}-1$. I.e. for every $z>1$ we have

$$
f\left(k+\frac{1}{z}-1\right)=z
$$

We must have $k+\frac{1}{z}-1 \in \mathbb{R}^{+}$ for each $z>1$ and taking the limit as $z$ tends to infinity we deduce that $k \geqslant 1$. (Without mentioning limits, assuming for contradiction that $k<1$, we can find a $z>1$ such that $k+\frac{1}{z}-1 \leqslant 0$, which is a contradiction.) Therefore, $k \geqslant 1$.

For $x>1$, i.e. $f(x)=\frac{1}{x}$ for every $x>1$ consider $P(1, x-1)$ to get $f(f(x))=(x-1) f(1)+1=x=f\left(\frac{1}{x}\right)$. Injectivity of $f$ shows that $f(x)=\frac{1}{x}$.
So for all possible values of $x$ we have shown that $f(x)=\frac{1}{x}$.