Answer
0
Work Step by Step
We find:
$\nabla\times\textbf{F}=\begin{vmatrix}\textbf{i}&\textbf{j}&\textbf{k}\\\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\e^{xz}&3xe^{y}&-e^{yz}\end{vmatrix}$
$=(\frac{\partial (-e^{yz})}{\partial y}-\frac{\partial (3xe^{y})}{\partial z})\textbf{i}-(\frac{\partial(-e^{yz})}{\partial x}-\frac{\partial(e^{xz})}{\partial z})\textbf{j}+(\frac{\partial(3xe^{y})}{\partial x}-\frac{\partial(e^{xz})}{\partial y})\textbf{k}$
$=(-ze^{yz}-0)\textbf{i}-(0-xe^{xz})\textbf{j}+(3e^{y}-0)\textbf{k}$
$=-ze^{yz}\textbf{i}+xe^{xz}\textbf{j}+3e^{y}\textbf{k}$
$\nabla\cdot(\nabla\times\textbf{F})=\nabla\cdot\langle-ze^{yz},xe^{xz},3e^{y}\rangle$
$=\frac{\partial}{\partial x}(-ze^{yz})+\frac{\partial}{\partial y}(xe^{xz})+\frac{\partial}{\partial z}(3e^{y})$
$=0+0+0=0$
