1. **State the problem:** Simplify the expression $$\left(\frac{2x^{-3} y^{-2}}{3x^{-2} y^3}\right)^{-2}$$ and then evaluate it for $x = -2$ and $y = 3$. 2. **Use the laws of exponents:** When dividing powers with the same base, subtract the exponents: $$\frac{a^m}{a^n} = a^{m-n}$$. When raising a power to another power, multiply the exponents: $$(a^m)^n = a^{mn}$$. 3. **Simplify inside the parentheses first:** $$\frac{2x^{-3} y^{-2}}{3x^{-2} y^3} = \frac{2}{3} \cdot x^{-3 - (-2)} \cdot y^{-2 - 3} = \frac{2}{3} x^{-1} y^{-5}$$ 4. **Apply the outer exponent $-2$:** $$\left(\frac{2}{3} x^{-1} y^{-5}\right)^{-2} = \left(\frac{2}{3}\right)^{-2} \cdot (x^{-1})^{-2} \cdot (y^{-5})^{-2}$$ 5. **Simplify each term:** $$\left(\frac{2}{3}\right)^{-2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4}$$ $$ (x^{-1})^{-2} = x^{2}$$ $$ (y^{-5})^{-2} = y^{10}$$ 6. **Combine all:** $$\frac{9}{4} x^{2} y^{10}$$ 7. **Substitute $x = -2$ and $y = 3$:** $$\frac{9}{4} \cdot (-2)^2 \cdot 3^{10} = \frac{9}{4} \cdot 4 \cdot 59049$$ 8. **Cancel and multiply:** $$\frac{9}{\cancel{4}} \cdot \cancel{4} \cdot 59049 = 9 \cdot 59049 = 531441$$ **Final answer:** $$531441$$