1. The problem is to find the value of $9^{\frac{3}{4}}$. 2. Recall the rule for fractional exponents: $a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m$. 3. Here, $a=9$, $m=3$, and $n=4$. So, $9^{\frac{3}{4}} = \left(\sqrt[4]{9}\right)^3$. 4. Calculate $\sqrt[4]{9}$. Since $9 = 3^2$, $\sqrt[4]{9} = \sqrt[4]{3^2} = 3^{\frac{2}{4}} = 3^{\frac{1}{2}} = \sqrt{3}$. 5. Now raise $\sqrt{3}$ to the power 3: $\left(\sqrt{3}\right)^3 = \left(3^{\frac{1}{2}}\right)^3 = 3^{\frac{3}{2}}$. 6. Simplify $3^{\frac{3}{2}} = 3^{1 + \frac{1}{2}} = 3 \times \sqrt{3}$. 7. Therefore, $9^{\frac{3}{4}} = 3 \sqrt{3}$. Final answer: $9^{\frac{3}{4}} = 3 \sqrt{3}$.