1. **State the problem:** Determine if the vector field $$\vec{F} = \left( 4y^{2} + \frac{3x^{2}y}{z^{2}} \right) \vec{i} + \left( 8xy + \frac{x^{3}}{z^{2}} \right) \vec{j} + \left( 11 - \frac{2x^{3}y}{z^{3}} \right) \vec{k}$$ is conservative. 2. **Recall the definition:** A vector field $$\vec{F} = P\vec{i} + Q\vec{j} + R\vec{k}$$ is conservative if there exists a scalar potential function $$\phi$$ such that $$\vec{F} = \nabla \phi$$. A necessary condition for $$\vec{F}$$ to be conservative in a simply connected domain is that its curl is zero: $$\nabla \times \vec{F} = \vec{0}$$ 3. **Identify components:** $$P = 4y^{2} + \frac{3x^{2}y}{z^{2}}, \quad Q = 8xy + \frac{x^{3}}{z^{2}}, \quad R = 11 - \frac{2x^{3}y}{z^{3}}$$ 4. **Compute partial derivatives for curl:** $$\frac{\partial R}{\partial y} = \frac{\partial}{\partial y} \left( 11 - \frac{2x^{3}y}{z^{3}} \right) = - \frac{2x^{3}}{z^{3}}$$ $$\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z} \left( 8xy + \frac{x^{3}}{z^{2}} \right) = - \frac{2x^{3}}{z^{3}}$$ $$\frac{\partial P}{\partial z} = \frac{\partial}{\partial z} \left( 4y^{2} + \frac{3x^{2}y}{z^{2}} \right) = - \frac{6x^{2}y}{z^{3}}$$ $$\frac{\partial R}{\partial x} = \frac{\partial}{\partial x} \left( 11 - \frac{2x^{3}y}{z^{3}} \right) = - \frac{6x^{2}y}{z^{3}}$$ $$\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} \left( 8xy + \frac{x^{3}}{z^{2}} \right) = 8y + \frac{3x^{2}}{z^{2}}$$ $$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} \left( 4y^{2} + \frac{3x^{2}y}{z^{2}} \right) = 8y + \frac{3x^{2}}{z^{2}}$$ 5. **Calculate curl components:** $$\left( \nabla \times \vec{F} \right)_x = \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z} = - \frac{2x^{3}}{z^{3}} - \left(- \frac{2x^{3}}{z^{3}} \right) = 0$$ $$\left( \nabla \times \vec{F} \right)_y = \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x} = - \frac{6x^{2}y}{z^{3}} - \left(- \frac{6x^{2}y}{z^{3}} \right) = 0$$ $$\left( \nabla \times \vec{F} \right)_z = \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = \left(8y + \frac{3x^{2}}{z^{2}} \right) - \left(8y + \frac{3x^{2}}{z^{2}} \right) = 0$$ 6. **Conclusion:** Since $$\nabla \times \vec{F} = \vec{0}$$ everywhere in the domain where $$z \neq 0$$ (to avoid division by zero), the vector field $$\vec{F}$$ is conservative in any simply connected region excluding the plane $$z=0$$. **Final answer:** The vector field is conservative in its domain where $$z \neq 0$$.