1. **State the problem:** Differentiate the function $y = \sin^{-3}(3x)$ with respect to $x$. 2. **Rewrite the function:** $\sin^{-3}(3x)$ means $\left(\sin(3x)\right)^{-3}$. 3. **Use the chain rule:** If $y = [u(x)]^n$, then $\frac{dy}{dx} = n[u(x)]^{n-1} \cdot u'(x)$. 4. **Apply the rule:** Here, $u(x) = \sin(3x)$ and $n = -3$. 5. **Find $u'(x)$:** $\frac{d}{dx} \sin(3x) = 3 \cos(3x)$. 6. **Combine:** $$\frac{dy}{dx} = -3 \left(\sin(3x)\right)^{-4} \cdot 3 \cos(3x) = -9 \frac{\cos(3x)}{\sin^{4}(3x)}$$ 7. **Final answer:** $$\frac{d}{dx} \sin^{-3}(3x) = -9 \frac{\cos(3x)}{\sin^{4}(3x)}$$ **Note:** The expression $2 \sin 3x \cos 3x$ is not the derivative of $\sin^{-3}(3x)$; it resembles the derivative of $\sin^2(3x)$.