1. **State the problem:** Evaluate the expression $$\frac{x^{-2}}{y} = \left(\frac{x}{y}\right)^2$$ when $$x=3$$ and $$y=-4$$. 2. **Recall the rules:** - Negative exponents mean reciprocal: $$x^{-2} = \frac{1}{x^2}$$. - When dividing by a variable, treat it as multiplication by its reciprocal. - Squaring a fraction means squaring numerator and denominator separately. 3. **Evaluate the left side:** $$\frac{x^{-2}}{y} = \frac{\frac{1}{x^2}}{y} = \frac{1}{x^2 \cdot y}$$ Substitute $$x=3$$ and $$y=-4$$: $$= \frac{1}{3^2 \cdot (-4)} = \frac{1}{9 \cdot (-4)} = \frac{1}{-36} = -\frac{1}{36}$$ 4. **Evaluate the right side:** $$\left(\frac{x}{y}\right)^2 = \left(\frac{3}{-4}\right)^2 = \frac{3^2}{(-4)^2} = \frac{9}{16}$$ 5. **Check equality:** Left side is $$-\frac{1}{36}$$ and right side is $$\frac{9}{16}$$, so they are not equal for these values. 6. **Final answer:** The value of the left side is $$-\frac{1}{36}$$ and the right side is $$\frac{9}{16}$$ when $$x=3$$ and $$y=-4$$. Hence, $$\frac{x^{-2}}{y} = -\frac{1}{36}$$ and $$\left(\frac{x}{y}\right)^2 = \frac{9}{16}$$.