1. **Simplify** $\left(\frac{81}{16}\right)^{-\frac{3}{4}}$. 2. Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$. 3. Also, for fractional exponents: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$. 4. First, simplify inside the parentheses: $\frac{81}{16} = \left(\frac{3^4}{2^4}\right)$. 5. So, $\left(\frac{81}{16}\right)^{-\frac{3}{4}} = \left(\frac{3^4}{2^4}\right)^{-\frac{3}{4}}$. 6. Apply the exponent: $\left(\frac{3^4}{2^4}\right)^{-\frac{3}{4}} = \frac{1}{\left(\frac{3^4}{2^4}\right)^{\frac{3}{4}}}$. 7. Simplify the inner exponent: $\left(\frac{3^4}{2^4}\right)^{\frac{3}{4}} = \frac{3^{4 \times \frac{3}{4}}}{2^{4 \times \frac{3}{4}}} = \frac{3^3}{2^3} = \frac{27}{8}$. 8. So the expression becomes $\frac{1}{\frac{27}{8}} = \frac{8}{27}$. **Final answer:** $\boxed{\frac{8}{27}}$.