To determine whether the given vector field $\mathbf{F}(x, y) = (3y - x^2)\mathbf{i} + (3x + y)\mathbf{j}$ is conservative, we need to check the following conditions:

1. Test if the curl is zero: A vector field is conservative if its curl is zero in two dimensions. For a vector field $\mathbf{F}(x, y) = P(x, y)\mathbf{i} + Q(x, y)\mathbf{j}$, the curl condition is: $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}.$

   Here:

   • $P(x, y) = 3y - x^2$
   • $Q(x, y) = 3x + y$

   Compute the partial derivatives: $\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3x + y) = 3,$ $\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(3y - x^2) = 3.$

   Since $\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}$, the curl is zero, so the vector field is conservative.

2. Find the potential function: A potential function $f(x, y)$ satisfies: $\nabla f = \mathbf{F}.$

   This means: $\frac{\partial f}{\partial x} = P(x, y) = 3y - x^2,$ $\frac{\partial f}{\partial y} = Q(x, y) = 3x + y.$

   Integrate $\frac{\partial f}{\partial x} = 3y - x^2$ with respect to $x$: $f(x, y) = \int (3y - x^2) \, dx = 3yx - \frac{x^3}{3} + h(y),$ where $h(y)$ is an arbitrary function of $y$.

   Differentiate this with respect to $y$: [ \frac{\