1. The problem is to find the derivative $\frac{dy}{dx}$ of the function $y = \sin(2x^3 + 3)$.\n\n2. We use the chain rule for differentiation, which states that if $y = \sin(u)$ and $u$ is a function of $x$, then $\frac{dy}{dx} = \cos(u) \cdot \frac{du}{dx}$.\n\n3. Here, $u = 2x^3 + 3$. We first find $\frac{du}{dx}$.\n\n4. Differentiating $u$ with respect to $x$, we get $\frac{du}{dx} = 6x^2$ because the derivative of $2x^3$ is $6x^2$ and the derivative of a constant $3$ is $0$.\n\n5. Applying the chain rule, $\frac{dy}{dx} = \cos(2x^3 + 3) \cdot 6x^2$.\n\n6. Therefore, the derivative is $$\frac{dy}{dx} = 6x^2 \cos(2x^3 + 3).$$