Curl $\bar{F} = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ r^{-3}x & r^{-3}y & r^{-3}z \end{vmatrix} \\ = i \bigg[\frac{\partial}{\partial y}(0) - \frac{\partial}{\partial z}(y^2 + x^2 y)\bigg] - j\bigg[\frac{\partial}{\partial x}(0) - \frac{\partial}{\partial z}(x^2 + xy^2)\bigg] + k\bigg[\frac{\partial}{\partial x}(y^2 + x^2 y) - \frac{\partial}{\partial y}(x^2 + xy^2)\bigg] \\ = i[0 - 0] - j[0 - 0] + k[2xy - 2xy] \\ = 0$

$\therefore \bar{F}$ is irrotational.

Now, if $\phi$ is the scalar potential then $\bar{F} = \bar{V}\phi$

$\therefore (x^2 + xy^2)i + (y^2 + x^2 y)j + 0k = \frac{\partial \phi}{\partial x}i + \frac{\partial \phi}{\partial y}j + \frac{\partial \phi}{\partial x}k \\ \therefore \frac{\partial \phi}{\partial x} = x^2 + xy^2, \frac{\partial \phi}{\partial y} = y^2 + x^2 y, \frac{\partial \phi}{\partial z} = 0$

But $\partial \phi = \frac{\partial \phi}{\partial x}dx + \frac{\partial \phi}{\partial y}dy + \frac{\partial \phi}{\partial z}dz \\ \partial \phi = (x^2 + xy^2)dx + (y^2 + x^2 y)dy + 0dz$

By Integration

$\phi = \frac{x^3}{3} + \frac{x^2 y^2}{2} + \frac{y^3}{3} + \frac{x^2 y^2}{2} \\ \phi = \frac{x^3}{3} + \frac{y^3}{3} + \frac{x^2 y^2}{2}$