Solve the equation $\log_3 x - 3\log_x 9 = 1$.

1. Change the base of the second logarithmic term to base 3

$\log_x 9 = \frac{\log_3 9}{\log_3 x} = \frac{2}{\log_3 x}$

2. Substitute into the equation and form a quadratic equation

Let $y = \log_3 x$.
$y - 3\left(\frac{2}{y}\right) = 1$
$y - \frac{6}{y} = 1$
$y^2 - y - 6 = 0$

3. Solve the quadratic equation

$(y - 3)(y + 2) = 0$
$y = 3$ or $y = -2$

4. Solve for $x$ using the definition of logarithms

When $y = 3$:
$\log_3 x = 3 \implies x = 3^3 = 27$

When $y = -2$:
$\log_3 x = -2 \implies x = 3^{-2} = \frac{1}{9}$