1. **State the problem:** Evaluate $$\log_{\frac{1}{9}} \left( \sqrt[3]{81} \right)^2$$. 2. **Rewrite the expression inside the logarithm:** $$\left( \sqrt[3]{81} \right)^2 = \left(81^{\frac{1}{3}}\right)^2 = 81^{\frac{2}{3}}$$. 3. **Express 81 as a power of 3:** $$81 = 3^4$$ So, $$81^{\frac{2}{3}} = \left(3^4\right)^{\frac{2}{3}} = 3^{4 \times \frac{2}{3}} = 3^{\frac{8}{3}}$$. 4. **Rewrite the base of the logarithm:** $$\frac{1}{9} = 9^{-1} = (3^2)^{-1} = 3^{-2}$$. 5. **Use the change of base formula for logarithms:** $$\log_{3^{-2}} 3^{\frac{8}{3}} = \frac{\log_3 3^{\frac{8}{3}}}{\log_3 3^{-2}}$$. 6. **Simplify the numerator and denominator using the power rule of logarithms:** $$\log_3 3^{\frac{8}{3}} = \frac{8}{3}$$ $$\log_3 3^{-2} = -2$$ 7. **Calculate the value:** $$\frac{\frac{8}{3}}{-2} = \frac{8}{3} \times \frac{1}{-2} = -\frac{8}{6} = -\frac{4}{3}$$. **Final answer:** $$-\frac{4}{3}$$ which corresponds to option d.