Given
[tex]\vec F = (4xz+y^2)\,\vec\imath + 2xy\,\vec\jmath + 2x^2\,\vec k[/tex]
its curl would be
[tex]\nabla\times\vec F = \left(\dfrac{\partial(2xy)}{\partial z} - \dfrac{\partial(2x^2)}{\partial y}\right)\,\vec\imath - \left(\dfrac{\partial(4xz+y^2)}{\partial z} - \dfrac{\partial(2x^2)}{\partial x}\right)\,\vec\jmath + \left(\dfrac{\partial(4xz+y^2)}{\partial y} - \dfrac{\partial(2xy)}{\partial x}\right)\,\vec k[/tex]
which reduces to the zero vector. Since the curl is zero, and [tex]\vec F[/tex] doesn't have any singularities, [tex]\vec F[/tex] is indeed a gradient field.
To determine what it is a gradient of, we look for a scalar function f(x, y, z) such that [tex]\nabla f = \vec F[/tex]. This entails solving for f such that
[tex]\dfrac{\partial f}{\partial x} = 4xz+y^2 \\\\ \dfrac{\partial f}{\partial y} = 2xy \\\\ \dfrac{\partial f}{\partial z} = 2x^2[/tex]
Integrate both sides of the first equation with respect to x, which gives
[tex]f(x,y,z) = 2x^2z+xy^2 + g(y,z)[/tex]
Differentiate both sides of this with respect to y :
[tex]\dfrac{\partial f}{\partial y} = 2xy = 2xy+\dfrac{\partial g}{\partial y} \\\\ \implies \dfrac{\partial g}{\partial y} = 0 \implies g(y,z) = h(z)[/tex]
Differentiate f with respect to z :
[tex]\dfrac{\partial f}{\partial z} = 2x^2 = 2x^2 + \dfrac{\mathrm dh}{\mathrm dz} \\\\ \implies \dfrac{\mathrm dh}{\mathrm dz} = 0\implies h(z) = C[/tex]
So it turns out that [tex]\vec F[/tex] is the gradient of
[tex]f(x,y,z) = 2x^2z+xy^2+C[/tex]
