Answer
\begin{align*}
\mathbf{div(F)}&= \frac{-y}{x^2}+ \frac{1}{z}+ \frac{1}{x} \\
\text{curl} \mathbf{(F)}&= \left( \frac{y}{z^2} \right)i - \left( \frac{-z}{x^2} \right)j + \left(\frac{-1}{x} \right) k
\end{align*}
Work Step by Step
Given $$\mathbf{F}=\left\langle\frac{y}{x}, \frac{y}{z}, \frac{z}{x}\right\rangle$$
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{y}{x})}{\partial x}+\frac{\partial (\frac{y}{z}) }{\partial y}+\frac{\partial ( \frac{z}{x})}{\partial z}\\
&= \frac{-y}{x^2}+ \frac{1}{z}+ \frac{1}{x}
\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{y}{x}& \frac{y}{z}& \frac{z}{x}
\end{array} \\
&= \left( \frac{\partial( \frac{z}{x})}{\partial y}- \frac{\partial(\frac{y}{z})}{\partial z}\right)i - \left( \frac{\partial( \frac{z}{x} )}{\partial x}- \frac{\partial(\frac{y}{x})}{\partial z}\right)j + \left( \frac{\partial( \frac{y}{z})}{\partial x}- \frac{\partial( \frac{y}{x} )}{\partial y}\right) k\\
&= \left( \frac{y}{z^2} \right)i - \left( \frac{-z}{x^2} \right)j + \left(\frac{-1}{x} \right) k\\
\end{align*}
