Answer
\begin{align*}
\mathbf{div(F)}& =3\\
\text{curl} \mathbf{(F)}&=0
\end{align*}
Work Step by Step
Given $$x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$$
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 (x)}{\partial x}+\frac{\partial (y) }{\partial y}+\frac{\partial (z)}{\partial z}\\
&=1+1+1=3
\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}\\
x& y& z
\end{array} \\
&= \left( \frac{\partial( z )}{\partial y}- \frac{\partial(y)}{\partial z}\right)i - \left( \frac{\partial( z )}{\partial x}- \frac{\partial(x )}{\partial z}\right)j + \left( \frac{\partial(y )}{\partial x}- \frac{\partial(x )}{\partial y}\right) k\\
&=0
\end{align*}
