Answer
\begin{align*}
\mathbf{div(F)} &=0\\
\text{curl} \mathbf{(F)} &=(\sin x) j + \left(\cos x-e^y \right) k
\end{align*}
Work Step by Step
Given $$\mathbf{F}=\left\langle e^{y}, \sin x, \cos x\right\rangle$$
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 ( e^{y})}{\partial x}+\frac{\partial (\sin x) }{\partial y}+\frac{\partial ( \cos x)}{\partial z}\\
&=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}\\
e^{y}& \sin x& \frac{z}{x}
\end{array} \\
&= \left( \frac{\partial( \cos x)}{\partial y}- \frac{\partial(\sin x)}{\partial z}\right)i - \left( \frac{\partial( \cos x )}{\partial x}- \frac{\partial(e^{y})}{\partial z}\right)j + \left( \frac{\partial(\sin x)}{\partial x}- \frac{\partial( e^{y} )}{\partial y}\right) k\\
&= \left( 0 \right)i - \left( -\sin x \right)j + \left(\cos x-e^y \right) k\\
&=(\sin x) j + \left(\cos x-e^y \right) k
\end{align*}
