Answer
$ \boxed{
\operatorname{div} \mathbf{F} = \frac{1}{x} + xz e^{xyz} + \frac{x}{x^2+z^2},
\quad
\nabla \times \mathbf{F} = -xy e^{xyz} \,\mathbf{i} + \frac{z}{x^2+z^2} \,\mathbf{j} + yz e^{xyz} \,\mathbf{k}
} $
Work Step by Step
Let
$\mathbf{F}(x, y, z) = \ln x \,\mathbf{i} + e^{xyz} \,\mathbf{j} + \tan^{-1}\left(\frac{z}{x}\right) \,\mathbf{k}.$
Divergence of $\mathbf{F}$
The divergence is defined as
$\operatorname{div} \mathbf{F} = \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}.$
Here,
$F_x = \ln x, \quad F_y = e^{xyz}, \quad F_z = \tan^{-1}\left(\frac{z}{x}\right).$
Compute each derivative:
$\frac{\partial F_x}{\partial x} = \frac{d}{dx} (\ln x) = \frac{1}{x},$
$\frac{\partial F_y}{\partial y} = \frac{\partial}{\partial y} \left(e^{xyz}\right) = xz e^{xyz},$
$\frac{\partial F_z}{\partial z} = \frac{\partial}{\partial z} \left(\tan^{-1}\frac{z}{x}\right) = \frac{1}{1+(z/x)^2} \cdot \frac{1}{x} = \frac{x}{x^2+z^2}.$
Therefore,
$\operatorname{div} \mathbf{F} = \frac{1}{x} + xz e^{xyz} + \frac{x}{x^2+z^2}.$
Curl of $\mathbf{F}$
The curl is defined as
$\nabla \times \mathbf{F} =$
$\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_x & F_y & F_z
\end{vmatrix}.$
Compute each component:
$(\nabla \times \mathbf{F})_x = \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}
= 0 - xy e^{xyz} = -xy e^{xyz},$
$(\nabla \times \mathbf{F})_y = \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}
= 0 - \frac{\partial}{\partial x} \left(\tan^{-1}\frac{z}{x}\right)
= - \frac{-z/x^2}{1+(z/x)^2} = \frac{z}{x^2+z^2},$
$
(\nabla \times \mathbf{F})_z = \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}
= \frac{\partial}{\partial x} \left(e^{xyz}\right) - 0 = yz e^{xyz}.$
Thus,
$\nabla \times \mathbf{F} = -xy e^{xyz} \,\mathbf{i} + \frac{z}{x^2+z^2} \,\mathbf{j} + yz e^{xyz} \,\mathbf{k}.$
