Answer
We verify that ${\bf{F}} = \nabla f$.
$\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Work Step by Step
1. We have ${\bf{F}}\left( {x,y,z} \right) = \frac{z}{x}{\bf{i}} + {\bf{j}} + \ln x{\bf{k}}$ and $f\left( {x,y,z} \right) = y + z\ln x$.
Evaluate
$\nabla f = \left( {\frac{{\partial f}}{{\partial x}},\frac{{\partial f}}{{\partial y}},\frac{{\partial f}}{{\partial z}}} \right) = \left( {\frac{z}{x},1,\ln x} \right)$
Since ${\bf{F}}\left( {x,y,z} \right) = \frac{z}{x}{\bf{i}} + {\bf{j}} + \ln x{\bf{k}}$, so ${\bf{F}} = \nabla f$.
2. Since ${\bf{F}} = \nabla f$, so ${\bf{F}}$ is conservative. By Theorem 1, the circulation of ${\bf{F}}$ around a closed circle is zero:
$\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
