For which $x$ does the following inequality hold?

$$
\left(\log _{x} 3\right) \cdot\left(\log _{3 x} 3\right) \geq \log _{9 x} 3
$$
$$
\left(\log _{x} 3\right) \cdot\left(\log _{3 x} 3\right) \geq \log _{9 x} 3
$$

Since the base of the logarithmic function can only be a positive number different from 1, (1) can only be satisfied for $x$ such that $x, 3 x, 9 x$ are such, that is,

$$
x>0, \quad \text { and } \quad x \neq 1, \quad x \neq \frac{1}{3}, \quad x \neq \frac{1}{9} \text {. }
$$

From the definition of the logarithm, it follows that if $\log _{a} b$ is defined and different from 0, then $\log _{b} a$ is also defined and equal to the reciprocal of the former. Using this, and introducing the notation

$$
y=\log _{3} x
$$

we find that (1) is equivalent to the inequality

$$
\frac{1}{y(y+1)} \geq \frac{1}{y+2}
$$

From this, by rearrangement, we get the inequality

$$
\frac{(y-\sqrt{2})(y+\sqrt{2})}{y(y+1)(y+2)} \leq 0
$$

The fraction on the left side of (4) is negative when the number of negative factors among them is odd, that is, when an odd number of the numbers

$$
\begin{array}{lllll}
-2, & -\sqrt{2}, & -1, & 0, & \sqrt{2}
\end{array}
$$

are greater than $y$. This happens if all of them are greater than $y$: $y \leq-2$, or if three of them are greater than $y$: $-\sqrt{2}<y<1$, and if only the largest of them is greater than $y$: $0<y<\sqrt{2}$. To these, we must add the values $y= \pm \sqrt{2}$, where the left side of (4) is equal to 0, thus we get all solutions of (4):

$$
y<-2 ; \quad-\sqrt{2} \leq y<-1 ; \quad 0<y \leq \sqrt{2}
$$

From this, based on (3), we get the solution of (1):

$$
0<x<3^{-2} ; \quad 3^{-\sqrt{2}} \leq x<3^{-1} ; \quad 1<x \leq 3^{\sqrt{2}}
$$

(Since we performed equivalent transformations, the result does not need to be verified.)