Determine all functions $f:(0, \infty) \rightarrow \mathbb{R}$ satisfying  
$$ \left(x+\frac{1}{x}\right) f(y)=f(x y)+f\left(\frac{y}{x}\right) $$  
for all $x, y>0$.  (South Korea)
Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\left(t^{2}\right)$, $f(a t)$ and $f\left(a^{2} t^{2}\right)$, the relation (1) provides a system of linear equations:
$$
\begin{array}{llll}
x=y=t: & \left(t+\frac{1}{t}\right) f(t) & =f\left(t^{2}\right)+f(1) \\
x=\frac{t}{a}, y=a t: & \left(\frac{t}{a}+\frac{a}{t}\right) f(a t) & =f\left(t^{2}\right)+f\left(a^{2}\right) \\
x=a^{2} t, y=t: & \left(a^{2} t+\frac{1}{a^{2} t}\right) f(t) & =f\left(a^{2} t^{2}\right)+f\left(\frac{1}{a^{2}}\right) \\
x=y=a t: & \left(a t+\frac{1}{a t}\right) f(a t) & =f\left(a^{2} t^{2}\right)+f(1)
\end{array}
$$
In order to eliminate $f\left(t^{2}\right)$, take the difference of (2a) and (2b); from (2c) and (2d) eliminate $f\left(a^{2} t^{2}\right)$; then by taking a linear combination, eliminate $f(a t)$ as well:
$$
\begin{gathered}
\left(t+\frac{1}{t}\right) f(t)-\left(\frac{t}{a}+\frac{a}{t}\right) f(a t)=f(1)-f\left(a^{2}\right) \text { and } \\
\left(a^{2} t+\frac{1}{a^{2} t}\right) f(t)-\left(a t+\frac{1}{a t}\right) f(a t)=f\left(1 / a^{2}\right)-f(1), \text { so } \\
\left(\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)\right) f(t) \\
=\left(a t+\frac{1}{a t}\right)\left(f(1)-f\left(a^{2}\right)\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(f\left(1 / a^{2}\right)-f(1)\right) .
\end{gathered}
$$
Notice that on the left-hand side, the coefficient of $f(t)$ is nonzero and does not depend on $t$:
$$
\left(a t+\frac{1}{a t}\right)\left(t+\frac{1}{t}\right)-\left(\frac{t}{a}+\frac{a}{t}\right)\left(a^{2} t+\frac{1}{a^{2} t}\right)=a+\frac{1}{a}-\left(a^{3}+\frac{1}{a^{3}}\right)<0 .
$$
After dividing by this fixed number, we get
$$
f(t)=C_{1} t+\frac{C_{2}}{t}
$$
where the numbers $C_{1}$ and $C_{2}$ are expressed in terms of $a, f(1), f\left(a^{2}\right)$ and $f\left(1 / a^{2}\right)$, and they do not depend on $t$. The functions of the form (3) satisfy the equation:
$$
\left(x+\frac{1}{x}\right) f(y)=\left(x+\frac{1}{x}\right)\left(C_{1} y+\frac{C_{2}}{y}\right)=\left(C_{1} x y+\frac{C_{2}}{x y}\right)+\left(C_{1} \frac{y}{x}+C_{2} \frac{x}{y}\right)=f(x y)+f\left(\frac{y}{x}\right) .
$$