Find the range of values of $x$ such that the graph $y = \ln \left( \frac{x-2}{x-3} \right)^2$ is decreasing.

1. Differentiate the function with respect to $x$.

Let $y = \ln \left( \frac{x-2}{x-3} \right)^2$.
Using the chain rule and quotient rule:
$\frac{dy}{dx} = \frac{1}{\left( \frac{x-2}{x-3} \right)^2} \cdot 2\left( \frac{x-2}{x-3} \right) \cdot \frac{d}{dx}\left( \frac{x-2}{x-3} \right)$
$\frac{dy}{dx} = \frac{x-3}{x-2} \cdot 2 \cdot \frac{(x-3)(1) - (x-2)(1)}{(x-3)^2}$
$\frac{dy}{dx} = \frac{2(x-3)}{x-2} \cdot \frac{-1}{(x-3)^2}$
$\frac{dy}{dx} = \frac{-2}{(x-2)(x-3)}$

2. Set up the inequality for a decreasing function.

For the graph to be decreasing, $\frac{dy}{dx} < 0$.
$\frac{-2}{(x-2)(x-3)} < 0$

3. Solve the inequality to find the range of $x$.

Since the numerator $-2$ is negative, the denominator must be positive for the fraction to be negative.
$(x-2)(x-3) > 0$
Solving this quadratic inequality gives:
$x < 2 \quad \text{or} \quad x > 3$