Let's determine the continuous functions $f:(0, \infty) \rightarrow(0, \infty)$ such that for any positive numbers $x, y$,

$$
f\left(\frac{1}{f(x y)}\right)=f(x) f(y)
$$
Rewriting the functional equation for the function $g(x)=\log f\left(e^{-x}\right)$

$$
g(g(x+y))=g(x)+g(y)
$$

Writing this for the numbers 0 and $x+y$ as well,

$$
g(x)+g(y)=g(g(x+y))=g(g(0+(x+y)))=g(0)+g(x+y)
$$

Introducing the function $h(x)=g(x)-g(0)$, (2) can be written as $h(x+y)=h(x)+h(y)$. The function $h$ is continuous and additive, so $h(x)=a x$ for some real number $a$. With the notation $g(0)=b$, $g(x)=a x+b$. Substituting this back into (1)

$$
\left(a^{2}-a\right)(x+y)+(a-1) b=0
$$

This is possible in two ways: $a=1$ and $b$ is any real number, or $a=b=0$. These are the solutions to equation (1).

Writing the solutions back in terms of $f$, in the first case

$$
f(x)=e^{-g(\log x)}=\frac{e^{-b}}{x}
$$

and in the second case

$$
f(x)=1
$$

Thus, the solutions to the functional equation are the constant 1 and the functions of the form $\frac{c}{x}$ $(c>0)$.

Note. The solution proceeds similarly even if we do not assume continuity, but the additive function $h$ cannot be written so simply. Instead of equation (3), we get that for any $x$

$$
h(h(x))-h(x)=b-h(b)
$$

From substituting $x=0$, we get $h(b)=b$ and $h(h(x))=h(x)$, that is, $h$ is a projection. It is easy to verify that these two conditions are also sufficient, that is, the solutions to (1) are the functions of the form $g(x)=h(x+c)$, where $h$ is any projection.