We are given the vector field $\mathbf{F}(x,y,z) = (x^2y + yz^2, x^3 + 4xz + 2, 2x + y^2z)$, and our goal is to find the potential function $\phi(x,y,z)$ such that:

$\mathbf{F} = \nabla \phi.$

This means that $\mathbf{F}$ is the gradient of $\phi$, or:

$\mathbf{F} = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right).$

This gives the following system of partial differential equations:

$\frac{\partial \phi}{\partial x} = x^2y + yz^2, \quad \frac{\partial \phi}{\partial y} = x^3 + 4xz + 2, \quad \frac{\partial \phi}{\partial z} = 2x + y^2z.$

Step 1: Integrating with respect to $x$

First, we integrate the first equation with respect to $x$:

$\frac{\partial \phi}{\partial x} = x^2y + yz^2.$

Integrating term by term:

$\phi(x, y, z) = \int (x^2y + yz^2) \, dx = \frac{x^3}{3}y + xyz^2 + g(y, z),$

where $g(y, z)$ is a function of $y$ and $z$ (because the derivative with respect to $x$ eliminates terms independent of $x$).

Step 2: Differentiating with respect to $y$

Next, we differentiate this expression with respect to $y$ and compare with the second equation:

$\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y} \left( \frac{x^3}{3}y + xyz^2 + g(y, z) \right) = \frac{x^3}{3} + xz^2 + \frac{\partial g(y, z)}{\partial y}.$

From the second equation, we know:

$\frac{\partial \phi}{\partial y} = x^3 + 4xz + 2.$

Equating both expressions for $\frac{\partial \phi}{\partial y}$:

$\frac{x^3}{3} + xz^2 + \frac{\partial g(y, z)}{\partial y} = x^3 + 4xz + 2.$

Simplifying this:

$\frac{2x^3}{3} + xz^2 + \frac{\partial g(y, z)}{\partial y} = 4xz + 2.$

This gives:

$\frac{\partial g(y, z)}{\partial y} = 4xz - xz^2 - \frac{2x^3}{3} + 2.$

However, upon closer inspection of constants,