1. **Problem:** Find the divergence of the vector field $\vec{F} = x^{2} y \vec{i} - (z^{3} - 3x) \vec{j} + 4 y^{2} \vec{k}$. 2. **Formula:** The divergence of a vector field $\vec{F} = P \vec{i} + Q \vec{j} + R \vec{k}$ is given by: $$\nabla \cdot \vec{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}$$ 3. **Identify components:** - $P = x^{2} y$ - $Q = -(z^{3} - 3x) = -z^{3} + 3x$ - $R = 4 y^{2}$ 4. **Calculate partial derivatives:** - $\frac{\partial P}{\partial x} = \frac{\partial}{\partial x} (x^{2} y) = 2 x y$ - $\frac{\partial Q}{\partial y} = \frac{\partial}{\partial y} (-z^{3} + 3x) = 0$ since $z$ and $x$ are independent of $y$ - $\frac{\partial R}{\partial z} = \frac{\partial}{\partial z} (4 y^{2}) = 0$ since $y$ is independent of $z$ 5. **Sum the partial derivatives:** $$\nabla \cdot \vec{F} = 2 x y + 0 + 0 = 2 x y$$ 6. **Answer:** The divergence of $\vec{F}$ is $$\boxed{2 x y}$$