Given $\bar{F} = (x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k}$ is irrotational

Curl F = 0

$\bar{V} \times \bar{F} = \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ (x^2 - yz) & (y^2 - zx) & (z^2 - xy) \end{vmatrix}$

Now, here F is irrotational there exists a scalar potential $\phi$ such that $\bar{F} = \bar{V}\phi$

$\therefore (x^2 - yz)\hat{i} + (y^2 - zx)\hat{j} + (z^2 - xy)\hat{k} = \frac{\partial \phi}{\partial x}\hat{i} + \frac{\partial \phi}{\partial y}\hat{j} + \frac{\partial \phi}{\partial z}\hat{k} \\ \therefore \frac{\partial \phi}{\partial x} = (x^2 - yz), \frac{\partial \phi}{\partial y} = (y^2 - zx), \frac{\partial \phi}{\partial z} = (z^2 - xdy)$

Now

$d \phi = \frac{\partial \phi}{\partial x}\partial x + \frac{\partial \phi}{\partial y}\partial y + \frac{\partial \phi}{\partial z}\partial z = (x^2 - yz)dx + (y^2 - zx)dy + (z^2 - xdy)dz \\ d\phi = x^2 dx + y^2 dy + z^2 dz - (yzdx + xzdy + xydz) \\ d\phi = \phi\bigg[\frac{x^3}{3} + \frac{y^3}{3} + \frac{z^3}{3} - xyz\bigg] \\ \therefore \phi = \frac{x^3}{3} + \frac{y^3}{3} + \frac{z^3}{3} - xyz$