By the comment under Table 9.7.2, the curl of F can be computed by augmenting F with a zero third-component. The vanishing of this curl is then equivalent to the identity $f_y\=g_x$, where $\mathbf{F}\left(x\,y\right)\=f\left(x\,y\right) \mathbf{i}\+g\left(x\,y\right) \mathbf{j}$. An implementation of this approach is

$\frac{\partial }{\partial y}\left(2 x y-y^3\right)\= 2 x-3 y^2\=\frac{\partial }{\partial x}\left(x^2-3 x y^2\right)$ 

 A scalar potential for F is $u\=x y\left(x-y^2\right)$ and $\nabla u\=\left[\begin{array}{c}2 x y-y^3\\ x^2-3 x y^2\end{array}\right]\=\mathbf{F}$. Either way, it should be clear that F is conservative.

Table 9.7.2(a) provides a solution based on the vanishing of the curl of F, where F is augmented to have a zero third component.

It is also possible to exhibit a scalar potential whose gradient is the vector F. This is done in Table 9.7.2(b).

Table 9.7.2(c) provides a solution via an application of the Curl command.

Table 9.7.2(d) obtains a scalar potential and shows that its gradient is F.

The Student VectorCalculus package tries to infer coordinate systems and coordinate-variable names whenever possible. In Table 9.7.2(d), the Gradient command infers the coordinate variables from the scalar $u$. Since $u$ has only the two variables $x$ and $y$, the Gradient command assumes it should return a two-component vector.