Answer
We prove that
(a) ${\bf{F}}$ is path-independent.
(b) ${\bf{F}}$ is conservative.
Work Step by Step
(a) Let ${{\cal C}_1}$ and ${{\cal C}_2}$ be any two paths in ${\cal D}$ with the same initial and terminal points $A$ and $B$. Thus, $ {{\cal C}_1}$ and $ - {{\cal C}_2}$ constitute a closed path ${\cal C}$ in ${\cal D}$, where the minus sign implies the reverse direction, that is from $B$ to $A$.
Denote the integral from $A$ to $B$ along the path ${{\cal C}_1}$ by $\mathop \smallint \limits_{{{\cal C}_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$. Likewise, denote the integral from $A$ to $B$ along the path ${{\cal C}_2}$ by $\mathop \smallint \limits_{{{\cal C}_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$.
Since $\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$, so
$\mathop \oint \limits_{\cal C} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{{\cal C}_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} - \mathop \smallint \limits_{{{\cal C}_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 0$
Therefore, $\mathop \smallint \limits_{{{\cal C}_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{{\cal C}_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$.
In other words, ${\bf{F}}$ is path-independent.
(b) From the result in part (a), we get $\mathop \smallint \limits_{{{\cal C}_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_{{{\cal C}_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$, where ${{\cal C}_1}$ and ${{\cal C}_2}$ are any two paths in ${\cal D}$ with the same initial and terminal points $A$ and $B$. This implies that ${\bf{F}}$ is path-independent. By Theorem 2, ${\bf{F}}$ is conservative.
