Example 2. Solve the equation

$$
\frac{|\operatorname{ctg} x y|}{\cos ^{2} x y}-2=\log _{\frac{1}{3}}\left(9 y^{2}-18 y+10\right)
$$
Solution. The methods considered in Example 1 cannot be used here. Let's try to use the property of boundedness of functions.

On the one hand,

$$
\frac{|\operatorname{ctg} x y|}{\cos ^{2} x y}=\left|\frac{\operatorname{ctg} x y}{\cos ^{2} x y}\right|=\left|\frac{2}{\sin 2 x y}\right| \geqslant 2
$$

and accordingly $\frac{|\operatorname{ctg} x y|}{\cos ^{2} x y}-2 \geqslant 0$.

On the other hand,

$$
\log _{\frac{1}{3}}\left(9 y^{2}-18 y+10\right)=\log _{\frac{1}{3}}\left(9(y-1)^{2}+1\right) \leqslant 0
$$

since $9(y-1)^{2}+1 \geqslant 1$ for any values of $y$.

Since the left side of the equation is non-negative for $x, y \in \mathbf{R}$, and the right side is no more than zero, for the equation to have a solution, the following system must have a solution:

$$
\left\{\begin{array}{l}
\left|\frac{2}{\sin 2 x y}\right|=2 \\
\log _{\frac{1}{3}}\left(9 y^{2}-18 y+10\right)=0
\end{array}\right.
$$

which is equivalent to

$$
\left\{\begin{array}{l}
|\sin 2 x y|=1 \\
9 y^{2}-18 y+10=1
\end{array}\right.
$$

Solving this, we get: $x=\frac{\pi}{4}+\frac{\pi k}{2}, y=1, k \in \boldsymbol{Z}$.