$\mathbf{F}=\mathbf{F}(x,y)=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}}$ is the gradient of $f(x,y) =\tan ^{-1}\frac{y}{x},$ $x\neq 0,$ since $$\begin{equation*} \nabla\! \ f=\frac{\partial }{\partial x}\tan ^{-1}\frac{y}{x}\mathbf{i}+\frac{\partial }{\partial y}\tan ^{-1}\frac{y}{x}\mathbf{j}=\frac{ \frac{-y}{x^{2}}}{1+\frac{y^{2}}{x^2}}\mathbf{i}+\frac{\frac{1}{x}}{1+\frac{y^{2}}{x^2}}\mathbf{j}=\frac{-y\mathbf{i}+x\mathbf{j}}{x^{2}+y^{2}} \end{equation*}$$

So, $\mathbf{F}$ is a conservative vector field on any region $R$ that contains no points on the $y$-axis ($x=0$). Then by the corollary $$\begin{equation*} \int_{C}\mathbf{F}\,{\cdot}\, d\mathbf{r}=0 \end{equation*}$$

along any closed, piecewise-smooth curve $C$ that does not cross or touch the $y$-axis.