6. Let $\mathbb{R}^{+}$ denote the set of all positive real numbers. Determine all functions $f$ : $\mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that for all $x, y \in \mathbb{R}^{+}$,

$$
f(x) f(y)=f(y) f(x f(y))+\frac{1}{x y} .
$$

(Pavel Calábek)
Solution. We will show that the only function $f$ that satisfies the conditions of the problem is

$$
f(x)=1+\frac{1}{x}
$$

From the problem statement, it follows that $f(y) \neq 0$ for any $y>0$, thus

$$
f(x f(y))=f(x)-\frac{1}{x y f(y)}
$$

Let $f(1)=a>0$. By choosing $x=1$, and $y=1$ in equation (1), we successively obtain

$$
\begin{gathered}
f(f(y))=f(1)-\frac{1}{y f(y)}=a-\frac{1}{y f(y)} \quad\left(y \in \mathbb{R}^{+}\right) \\
f(a x)=f(x)-\frac{1}{a x} \quad\left(x \in \mathbb{R}^{+}\right) .
\end{gathered}
$$

By choosing $x=1$ in equation (3), we get

$$
f(a)=f(1)-\frac{1}{a}=a-\frac{1}{a} .
$$

By choosing $x=a$ in equation (1) and using (4), we get

$$
f(a f(y))=f(a)-\frac{1}{a y f(y)}=a-\frac{1}{a}-\frac{1}{a y f(y)} \quad\left(y \in \mathbb{R}^{+}\right),
$$

while using equations (3) and (2), the left side of the previous equation can be rewritten as

$$
f(a f(y))=f(f(y))-\frac{1}{a f(y)}=a-\frac{1}{y f(y)}-\frac{1}{a f(y)} .
$$

By comparing the right sides of the previous two equations, we compute

$$
f(y)=1+\frac{a-1}{y} \quad\left(y \in \mathbb{R}^{+}\right) .
$$

Therefore, if there is a solution to the given equation, it must have the form (4). By substituting into the equation in the problem statement and simplifying, we find that for all positive real $x$ and $y$, $(a-1)^{2}=1$ must hold. Given the assumption $a>0$, this equation is satisfied precisely when $a=2$. This step also verifies the correctness of the found solution.