Answer
We verify the identity:
$div\left( {curl\left( {\bf{F}} \right)} \right) = 0$
$div\left( {curl\left( {\bf{G}} \right)} \right) = 0$
Work Step by Step
1. For ${\bf{F}} = \left( {xz,y{{\rm{e}}^x},yz} \right)$.
$curl\left( {\bf{F}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{xz}&{y{{\rm{e}}^x}}&{yz}
\end{array}} \right|$
$curl\left( {\bf{F}} \right) = z{\bf{i}} + x{\bf{j}} + y{{\rm{e}}^x}{\bf{k}}$
$div\left( {curl\left( {\bf{F}} \right)} \right) = \frac{\partial }{{\partial x}}z + \frac{\partial }{{\partial y}}x + \frac{\partial }{{\partial z}}\left( {y{{\rm{e}}^x}} \right) = 0$
2. For ${\bf{G}} = \left( {{z^2},x{y^3},{x^2}y} \right)$.
$curl\left( {\bf{G}} \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{{z^2}}&{x{y^3}}&{{x^2}y}
\end{array}} \right|$
$curl\left( {\bf{G}} \right) = {x^2}{\bf{i}} - \left( {2xy - 2z} \right){\bf{j}} + {y^3}{\bf{k}}$
$div\left( {curl\left( {\bf{G}} \right)} \right) = \frac{\partial }{{\partial x}}{x^2} + \frac{\partial }{{\partial y}}\left( { - 2xy + 2z} \right) + \frac{\partial }{{\partial z}}{y^3} = 2x - 2x = 0$
