1. **State the problem:** Simplify the expression $$\left( \frac{3x^{\frac{3}{2}}y^3}{x^2 y^{-\frac{1}{2}}} \right)^{-2}$$. 2. **Apply the quotient rule for exponents:** When dividing like bases, subtract the exponents. Simplify inside the parentheses first: $$\frac{3x^{\frac{3}{2}}y^3}{x^2 y^{-\frac{1}{2}}} = 3 \cdot x^{\frac{3}{2} - 2} \cdot y^{3 - (-\frac{1}{2})} = 3 \cdot x^{-\frac{1}{2}} \cdot y^{3 + \frac{1}{2}} = 3x^{-\frac{1}{2}} y^{\frac{7}{2}}$$ 3. **Rewrite the expression:** $$\left(3x^{-\frac{1}{2}} y^{\frac{7}{2}}\right)^{-2}$$ 4. **Apply the power of a product rule:** Raise each factor to the power of -2. $$3^{-2} \cdot \left(x^{-\frac{1}{2}}\right)^{-2} \cdot \left(y^{\frac{7}{2}}\right)^{-2}$$ 5. **Simplify each term:** - $$3^{-2} = \frac{1}{3^2} = \frac{1}{9}$$ - $$\left(x^{-\frac{1}{2}}\right)^{-2} = x^{(-\frac{1}{2})(-2)} = x^1 = x$$ - $$\left(y^{\frac{7}{2}}\right)^{-2} = y^{\frac{7}{2} \times (-2)} = y^{-7} = \frac{1}{y^7}$$ 6. **Combine all simplified terms:** $$\frac{1}{9} \cdot x \cdot \frac{1}{y^7} = \frac{x}{9 y^7}$$ **Final answer:** $$\frac{x}{9 y^7}$$