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

$$
x^{2}(f(x)+f(y))=(x+y) f(f(x) y) \text {. }
$$

(P. Kaňovský)
Solution. Let $f$ be any of the sought functions. Denote $f(1)=p$, according to the conditions of the problem, $p>0$.

In the given relation, set $x=1, y=1$. After rearrangement, we get

$$
p=f(p) \text {. }
$$

In the given relation, further set $x=p, y=1$. Then

$$
p^{2}(f(p)+p)=(p+1) f(f(p))
$$

and according to (1), we get

$$
2 p^{3}=(p+1) p \text {. }
$$

This algebraic equation has three real roots $-\frac{1}{2}, 0,1$. The only root satisfying the condition $p>0$ is $p=1$, hence

$$
f(1)=1
$$

Let $t$ be any positive real number. In the given relation, set $x=1, y=t$, so according to (2) we get

$$
1+f(t)=(1+t) f(t) .
$$

From this, after rearrangement,

$$
f(t)=\frac{1}{t}
$$

By substitution, we easily verify that the function $f(t)=1 / t$ satisfies the equation from the problem statement. The function determined by relation (3) is the only solution to the given problem.