Answer
\begin{align*}
\mathbf{div(F)}&= 0\\
\text{curl} \mathbf{(F)}&=0
\end{align*}
Work Step by Step
Given $$\mathbf{F}=\left\langle\frac{x}{x^{2}+y^{2}}, \frac{y}{x^{2}+y^{2}}, 0\right\rangle$$
Since
\begin{align*}
\mathbf{div(F)}&=\frac{\partial \mathbf{F}_x}{\partial x}+\frac{\partial \mathbf{F}_y }{\partial y}+\frac{\partial \mathbf{F}_z}{\partial z}\\
&=\frac{\partial ( \frac{x}{x^{2}+y^{2}})}{\partial x}+\frac{\partial (\frac{y}{x^{2}+y^{2}}) }{\partial y}+\frac{\partial ( 0)}{\partial z}\\
&= \frac{-x^2+y^2}{\left(x^2+y^2\right)^2}+\frac{x^2-y^2}{\left(x^2+y^2\right)^2}\\
&=0
\end{align*}
and
\begin{align*}
\text{curl} \mathbf{(F)}&=\begin{array}{|||}
i&j &k\\
\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
\frac{x}{x^{2}+y^{2}}&\frac{y}{x^{2}+y^{2}}& 0
\end{array} \\
&= \left( \frac{\partial(0)}{\partial y}- \frac{\partial(\frac{y}{x^{2}+y^{2}})}{\partial z}\right)i - \left( \frac{\partial( 0 )}{\partial x}- \frac{\partial(\frac{x}{x^{2}+y^{2}})}{\partial z}\right)j +
\left( \frac{\partial(\frac{y}{x^{2}+y^{2}})}{\partial x}- \frac{\partial( \frac{x}{x^{2}+y^{2}} )}{\partial y}\right) k\\
&= \left(-\frac{2yx}{\left(x^2+y^2\right)^2}-\left(-\frac{2yx}{\left(x^2+y^2\right)^2}\right)\right) k\\
&=0
\end{align*}
