1. The problem is to evaluate $49^{-\frac{3}{2}}$. 2. Recall the rule for negative exponents: $a^{-n} = \frac{1}{a^n}$. 3. Also recall the fractional exponent rule: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$. 4. Apply the negative exponent rule: $$49^{-\frac{3}{2}} = \frac{1}{49^{\frac{3}{2}}}$$ 5. Now evaluate $49^{\frac{3}{2}}$ using the fractional exponent rule: $$49^{\frac{3}{2}} = \left(\sqrt{49}\right)^3$$ 6. Since $\sqrt{49} = 7$, substitute: $$\left(7\right)^3 = 7^3 = 343$$ 7. Substitute back into the fraction: $$\frac{1}{343}$$ 8. Therefore, the final answer is: $$49^{-\frac{3}{2}} = \frac{1}{343}$$