1. **State the problem:** Simplify the expression $$\frac{(3xy^{-2})^{-2}}{3x^{-2}y}$$. 2. **Recall the rules:** - Power of a product: $$(ab)^n = a^n b^n$$ - Negative exponent: $$a^{-n} = \frac{1}{a^n}$$ - Division of powers with the same base: $$\frac{a^m}{a^n} = a^{m-n}$$ 3. **Apply the power of a product rule to the numerator:** $$ (3xy^{-2})^{-2} = 3^{-2} \cdot x^{-2} \cdot (y^{-2})^{-2} $$ 4. **Simplify each term:** - $$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$$ - $$x^{-2} = \frac{1}{x^2}$$ - $$(y^{-2})^{-2} = y^{(-2) \times (-2)} = y^4$$ So numerator becomes: $$ \frac{1}{9} \cdot \frac{1}{x^2} \cdot y^4 = \frac{y^4}{9x^2} $$ 5. **Rewrite the denominator:** $$ 3x^{-2}y = 3 \cdot \frac{1}{x^2} \cdot y = \frac{3y}{x^2} $$ 6. **Write the full fraction:** $$ \frac{\frac{y^4}{9x^2}}{\frac{3y}{x^2}} $$ 7. **Divide the fractions by multiplying numerator by reciprocal of denominator:** $$ \frac{y^4}{9x^2} \times \frac{x^2}{3y} $$ 8. **Cancel common factors:** $$ \frac{y^4}{\cancel{9} \cancel{x^2}} \times \frac{\cancel{x^2}}{3y} = \frac{y^4}{9} \times \frac{1}{3y} $$ 9. **Multiply numerators and denominators:** $$ \frac{y^4 \times 1}{9 \times 3y} = \frac{y^4}{27y} $$ 10. **Simplify powers of $y$:** $$ \frac{y^4}{y} = y^{4-1} = y^3 $$ 11. **Final simplified expression:** $$ \frac{y^3}{27} $$ **Answer:** $$\boxed{\frac{y^3}{27}}$$