1. **State the problem:** Find the derivative $y'$ of the function $$y = \sin^4(x^3) = (\sin(x^3))^4.$$\n\n2. **Recall the chain rule:** If $y = [u(x)]^n$, then $$y' = n[u(x)]^{n-1} \cdot u'(x).$$\nAlso, for $u(x) = \sin(v(x))$, $$u'(x) = \cos(v(x)) \cdot v'(x).$$\n\n3. **Identify inner functions:** Here, $u(x) = \sin(x^3)$ and $v(x) = x^3$.\n\n4. **Differentiate step-by-step:**\n$$y' = 4(\sin(x^3))^{3} \cdot \frac{d}{dx}[\sin(x^3)]$$\n$$= 4(\sin(x^3))^{3} \cdot \cos(x^3) \cdot \frac{d}{dx}[x^3]$$\n$$= 4(\sin(x^3))^{3} \cdot \cos(x^3) \cdot 3x^{2}$$\n\n5. **Simplify the expression:**\n$$y' = 12x^{2} (\sin(x^3))^{3} \cos(x^3).$$\n\n**Final answer:** $$\boxed{y' = 12x^{2} \sin^{3}(x^{3}) \cos(x^{3})}.$$