$$ \frac { 4 } { 3 } \log _ { 3 } ^ { 2 } ( 5 x - 6 ) ^ { 3 } - \log _ { 3 } ( 5 x - 6 ) ^ { 3 } \cdot \log _ { 3 } x ^ { 6 } = - 6 \log _ { 3 } ^ { 2 } \frac { 1 } { x } $$

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To solve the equation:

$$
\frac{4}{3} \log_{3}^{2} \left( (5x - 6)^3 \right) - \log_{3} \left( (5x - 6)^3 \right) \cdot \log_{3} \left( x^6 \right) = -6 \log_{3}^{2} \left( \frac{1}{x} \right)
$$

let's break it down step by step.

### Step 1: Simplify the Equation

First, interpret the notation $$\log_{3}^{2}$$ as $$(\log_{3})^2$$. The equation becomes:

$$
\frac{4}{3} \left[\log_{3} \left( (5x - 6)^3 \right)\right]^2 - \log_{3} \left( (5x - 6)^3 \right) \cdot \log_{3} \left( x^6 \right) = -6 \left[\log_{3} \left( \frac{1}{x} \right)\right]^2
$$

Simplify each logarithmic expression:

$$
\log_{3} \left( (5x - 6)^3 \right) = 3 \log_{3} (5x - 6)
$$
$$
\log_{3} \left( x^6 \right) = 6 \log_{3} x
$$
$$
\log_{3} \left( \frac{1}{x} \right) = -\log_{3} x
$$

Substituting these back into the equation:

$$
\frac{4}{3} \left(3 \log_{3} (5x - 6)\right)^2 - 3 \log_{3} (5x - 6) \cdot 6 \log_{3} x = -6 \left(-\log_{3} x\right)^2
$$

Simplifying further:

$$
\frac{4}{3} \cdot 9 \log_{3}^2 (5x - 6) - 18 \log_{3} (5x - 6) \log_{3} x = -6 \log_{3}^2 x
$$

$$
12 \log_{3}^2 (5x - 6) - 18 \log_{3} (5x - 6) \log_{3} x + 6 \log_{3}^2 x = 0
$$

### Step 2: Introduce Substitutions

Let:

$$
a = \log_{3} (5x - 6) \quad \text{and} \quad b = \log_{3} x
$$

The equation becomes:

$$
12a^2 - 18ab + 6b^2 = 0
$$

Divide the entire equation by 6:

$$
2a^2 - 3ab + b^2 = 0
$$

This is a quadratic equation in terms of $$a$$:

$$
2a^2 - 3ab + b^2 = 0
$$

Solving for $$a$$:

$$
a = \frac{3b \pm \sqrt{(3b)^2 - 4 \cdot 2 \cdot b^2}}{4} = \frac{3b \pm b}{4}
$$

Thus, we have two cases:

1. **Case 1:** $$a = b$$
   
   $$
   \log_{3} (5x - 6) = \log_{3} x \implies 5x - 6 = x \implies 4x = 6 \implies x = \frac{3}{2}
   $$
   
2. **Case 2:** $$a = \frac{b}{2}$$
   
   $$
   \log_{3} (5x - 6) = \frac{1}{2} \log_{3} x \implies 5x - 6 = \sqrt{x}
   $$
   
   Let $$y = \sqrt{x}$$, then $$x = y^2$$:
   
   $$
   5y^2 - 6 = y \implies 5y^2 - y - 6 = 0
   $$
   
   Solving the quadratic equation:
   
   $$
   y = \frac{1 \pm \sqrt{1 + 120}}{10} = \frac{1 \pm 11}{10}
   $$
   
   Discarding the negative solution since $$y = \sqrt{x} > 0$$:
   
   $$
   y = \frac{12}{10} = 1.2 \implies x = (1.2)^2 = \frac{36}{25}
   $$

### Step 3: Verify the Solutions

1. **For $$x = \frac{3}{2}$$:**
   
   $$
   5\left(\frac{3}{2}\right) - 6 = \frac{15}{2} - 6 = \frac{3}{2} > 0
   $$
   
2. **For $$x = \frac{36}{25}$$:**
   
   $$
   5\left(\frac{36}{25}\right) - 6 = \frac{180}{25} - 6 = \frac{180 - 150}{25} = \frac{30}{25} = \frac{6}{5} > 0
   $$

Both solutions satisfy the original equation's domain restrictions.

### Final Answer

The solutions to the equation are:
$$
x = \frac{3}{2} \quad \text{and} \quad x = \frac{36}{25}
$$
The solutions are $$ x = \frac{3}{2} $$ and $$ x = \frac{36}{25} $$.

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