1. **State the problem:** Differentiate $\sin^3(3x^2 - 5)$ with respect to $x$. 2. **Recall the formula:** If $y = [f(x)]^n$, then by the chain rule, $\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)$. 3. **Apply the chain rule:** Here, $f(x) = \sin(3x^2 - 5)$ and $n=3$, so $$\frac{dy}{dx} = 3 \sin^2(3x^2 - 5) \cdot \frac{d}{dx}[\sin(3x^2 - 5)].$$ 4. **Differentiate the inner function:** Use the chain rule again for $\sin(3x^2 - 5)$: $$\frac{d}{dx}[\sin(3x^2 - 5)] = \cos(3x^2 - 5) \cdot \frac{d}{dx}[3x^2 - 5] = \cos(3x^2 - 5) \cdot 6x.$$ 5. **Combine all parts:** $$\frac{dy}{dx} = 3 \sin^2(3x^2 - 5) \cdot \cos(3x^2 - 5) \cdot 6x = 18x \sin^2(3x^2 - 5) \cos(3x^2 - 5).$$ **Final answer:** $$\boxed{\frac{d}{dx} \left( \sin^3(3x^2 - 5) \right) = 18x \sin^2(3x^2 - 5) \cos(3x^2 - 5)}.$$