Question

Consider the vector field f(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)kf(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)k.

Answer 1

$\dfrac{\partial f}{\partial x}=2z+3y\implies f(x,y,z)=2xz+3xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=3x+\dfrac{\partial g}{\partial y}=3z+3x$
$\dfrac{\partial g}{\partial y}=3z\implies g(y,z)=3yz+h(z)$
$\implies f(x,y,z)=3xz+3xy+3yz+h(z)$


$\dfrac{\partial f}{\partial z}=3x+3y+\dfrac{\mathrm dh}{\mathrm dz}=3y+2x$
$\dfrac{\mathrm dh}{\mathrm dz}=-x$

But we assume $h$ is a function of $z$ alone, so there is no solution for $h$ and hence no scalar function $f$ such that $\nabla f=\mathbf f$.