Example 39([31.4]) Let the set of all positive rational numbers be denoted as $Q^{+}$. Construct a function $f: Q^{+} \rightarrow Q^{+}$, such that for any $x, y \in Q^{+}$, it satisfies $y f(x f(y))=f(x)$.
Assume the function $f$ satisfies the given condition $y f(x f(y))=f(x)$. Let's first discuss the basic properties such a function $f$ should satisfy. Taking $x=1$, from the condition we get
$$f(f(y))=f(1) / y.$$

If $f(x)=f(y)$, then from this and equation (1) we derive
$$f(1) / y=f(f(y))=f(f(x))=f(1) / x, \text{ i.e., } x=y.$$

Therefore,
$f(x)=f(y)$ if and only if $x=y$.
Taking $x=y=1$, from the condition we get $f(f(1))=f(1)$. From this and equations (2), (1), we have
$$f(1)=1, \quad f(f(y))=1 / y.$$

Substituting $f(x)$ for $x$, from the condition and equation (3) we get
$$f(f(x) f(y))=f(f(x)) / y=1 /(x y)=f(f(x y)).$$

From this and equation (2), we derive
$$f(x y)=f(x) f(y),$$

i.e., $f(x)$ is a "totally multiplicative function". Since $f(x)$ is not equal to zero, it is easy to see (why) that the condition $y f(x f(y))=f(x)$ holds if and only if conditions (1) and (5) hold.

Now, the function $f$ is defined on $Q^{+}$. By the fundamental theorem of arithmetic, $x \in Q^{+}$ if and only if $x$ has the representation
$$x=p_{1}^{a(1)} \cdots p_{r}^{a(r)}, \quad a(j) \in \mathbb{Z}, 1 \leqslant j \leqslant r,$$

where $p_{1}, \cdots, p_{k}, \cdots$ are all prime numbers, arranged in increasing order. When $x$ is given by the above expression, from equation (5) and $f(1)=1$ we get
$$f(x)=f\left(p_{1}\right)^{a(1)} \cdots f\left(p_{r}\right)^{a(r)}.$$

Therefore, it is sufficient to define the value of the function $f$ at prime numbers $p$ to meet the required conditions. From the above analysis, it is known that it is sufficient for prime numbers $p$ to satisfy condition (1), i.e.,
$$f(f(p))=1 / p.$$

Since $f$ is a function from $Q^{+} \rightarrow Q^{+}$, we can assume
$$f(p)=q_{1}^{a(1)} \cdots q_{r}^{a(r)}, \quad q_{j} \text{ is a prime, } a(j) \in \mathbb{Z}, 1 \leqslant j \leqslant r.$$

From equation (5) we know
$$1 / p=f(f(p))=f\left(q_{1}\right)^{a(1)} \cdots f\left(q_{r}\right)^{a(r)}.$$

Since $f(1)=1$ only when $x=1$, the above equation suggests that $f(p)$ should take a very simple value, such as $q$ or $1 / q$, where $q$ is a prime. However, these discussions do not truly provide information on how to define the value of $f(p)$. Through experimentation and observation, for $j \geqslant 1$ we define
$$f\left(p_{2 j-1}\right)=1 / p_{2 j}, \quad f\left(p_{2 j}\right)=p_{2 j-1}$$

to meet the requirements.