The problem asks to simplify the following expression:

$\left( \frac{6x^3 y^{-3} z^{-4}}{9x^{-5} y z^{-3}} \right)^{-2}$

Step-by-Step Simplification

1. Simplify the expression inside the parentheses:

   Start by dividing the terms inside the fraction:

   $\frac{6x^3 y^{-3} z^{-4}}{9x^{-5} y z^{-3}} = \frac{6}{9} \cdot \frac{x^3}{x^{-5}} \cdot \frac{y^{-3}}{y} \cdot \frac{z^{-4}}{z^{-3}}$

2. Simplify constants: $\frac{6}{9} = \frac{2}{3}$

3. Simplify powers of $x$: Use the rule $\frac{a^m}{a^n} = a^{m-n}$: $\frac{x^3}{x^{-5}} = x^{3 - (-5)} = x^{3 + 5} = x^8$

4. Simplify powers of $y$: $\frac{y^{-3}}{y} = y^{-3 - 1} = y^{-4}$

5. Simplify powers of $z$: $\frac{z^{-4}}{z^{-3}} = z^{-4 - (-3)} = z^{-4 + 3} = z^{-1}$

   So, the expression inside the parentheses becomes: $\frac{2}{3} x^8 y^{-4} z^{-1}$

6. Apply the exponent of $-2$: Now apply the power $-2$ to the entire expression: $\left( \frac{2}{3} x^8 y^{-4} z^{-1} \right)^{-2}$

   Distribute the exponent $-2$ to each part: $\frac{2^{-2}}{3^{-2}} \cdot x^{8 \times (-2)} \cdot y^{-4 \times (-2)} \cdot z^{-1 \times (-2)}$

   Simplifying each part: $\frac{2^{-2}}{3^{-2}} = \frac{9}{4}, \quad x^{-16}, \quad y^8, \quad z^2$

   So, the final simplified expression is: $\frac{9}{4} x^{-16} y^8 z^2$

Or, alternatively: $\frac{9y^8z^2}{4x^{16}}$

This is the simplified form of the given expression.

Would you like further clarification or details on any step? Here are five related questions you might find helpful:

1. How can negative exponents be converted to positive ones?
2. What are the laws of exponents and how are they applied in algebraic expressions?
3. How do you handle fractional expressions with multiple variables?
4. What happens when an expression with negative exponents is raised to a negative power?
5. How would the process change if the exponent was a different value, like $-3$?

Tip: Always handle constants, variables, and exponents separately to avoid confusion when simplifying complex expressions.