1. The problem is to find the derivative of the function $f(x) = \sin(x^3)$.\n\n2. We use the chain rule for differentiation, which states that if $f(x) = \sin(g(x))$, then $f'(x) = \cos(g(x)) \cdot g'(x)$.\n\n3. Here, $g(x) = x^3$, so we first find $g'(x) = 3x^2$.\n\n4. Applying the chain rule, we get $$f'(x) = \cos(x^3) \cdot 3x^2 = 3x^2 \cos(x^3).$$\n\n5. Therefore, the derivative of $f(x) = \sin(x^3)$ is $f'(x) = 3x^2 \cos(x^3)$.