Answer
We prove: $div{\ }curl\left( {\bf{F}} \right) = 0$
Work Step by Step
Write the vector field: ${\bf{F}} = \left( {{F_1},{F_2},{F_3}} \right)$.
We assume that the partial derivatives exist and are continuous.
Evaluate the curl of ${\bf{F}}$:
$curl\left( {\bf{F}} \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{{F_1}}&{{F_2}}&{{F_3}}
\end{array}} \right)$
$curl\left( {\bf{F}} \right) = \left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right){\bf{i}} - \left( {\frac{{\partial {F_3}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial z}}} \right){\bf{j}} + \left( {\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right){\bf{k}}$
Evaluate the divergence of $curl\left( {\bf{F}} \right)$:
$divcurl\left( {\bf{F}} \right) = \left( {\frac{\partial }{{\partial x}},\frac{\partial }{{\partial y}},\frac{\partial }{{\partial z}}} \right)\cdot\left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}, - \frac{{\partial {F_3}}}{{\partial x}} + \frac{{\partial {F_1}}}{{\partial z}},\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)$
$ = \frac{\partial }{{\partial x}}\left( {\frac{{\partial {F_3}}}{{\partial y}} - \frac{{\partial {F_2}}}{{\partial z}}} \right) + \frac{\partial }{{\partial y}}\left( { - \frac{{\partial {F_3}}}{{\partial x}} + \frac{{\partial {F_1}}}{{\partial z}}} \right) + \frac{\partial }{{\partial z}}\left( {\frac{{\partial {F_2}}}{{\partial x}} - \frac{{\partial {F_1}}}{{\partial y}}} \right)$
$ = \frac{{{\partial ^2}{F_3}}}{{\partial x\partial y}} - \frac{{{\partial ^2}{F_2}}}{{\partial x\partial z}} - \frac{{{\partial ^2}{F_3}}}{{\partial y\partial x}} + \frac{{\partial {F_1}}}{{\partial y\partial z}} + \frac{{{\partial ^2}{F_2}}}{{\partial z\partial x}} - \frac{{{\partial ^2}{F_1}}}{{\partial z\partial y}}$
$ = \frac{{{\partial ^2}{F_3}}}{{\partial x\partial y}} - \frac{{{\partial ^2}{F_3}}}{{\partial y\partial x}} - \frac{{{\partial ^2}{F_2}}}{{\partial x\partial z}} + \frac{{{\partial ^2}{F_2}}}{{\partial z\partial x}} + \frac{{\partial {F_1}}}{{\partial y\partial z}} - \frac{{{\partial ^2}{F_1}}}{{\partial z\partial y}}$
Since $\frac{{{\partial ^2}{F_3}}}{{\partial x\partial y}} = \frac{{{\partial ^2}{F_3}}}{{\partial y\partial x}}{\rm{,}}\frac{{{\partial ^2}{F_2}}}{{\partial x\partial z}} = \frac{{{\partial ^2}{F_2}}}{{\partial z\partial x}}{\rm{,}}\frac{{\partial {F_1}}}{{\partial y\partial z}} = \frac{{{\partial ^2}{F_1}}}{{\partial z\partial y}}$, so
$div{\ }curl\left( {\bf{F}} \right) = 0$
