Answer
$div\left( {\bf{F}} \right) = y + z$
$curl\left( {\bf{F}} \right) = y{\bf{i}} + 3{x^2}{\bf{j}} - x{\bf{k}}$
Work Step by Step
We have ${\bf{F}} = \left( {xy,yz,{y^2} - {x^3}} \right)$.
Let ${F_1} = xy$, ${F_2} = yz$, ${F_3} = {y^2} - {x^3}$.
$div\left( {\bf{F}} \right) = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}} = y + z + 0$
So, $div\left( {\bf{F}} \right) = y + z$.
$curl\left( {\bf{F}} \right) = \left( {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\
{{F_1}}&{{F_2}}&{{F_3}}
\end{array}} \right)$
$ = \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}}$
$ = \left( {2y - y} \right){\bf{i}} - \left( { - 3{x^2} - 0} \right){\bf{j}} + \left( {0 - x} \right){\bf{k}}$
So, $curl\left( {\bf{F}} \right) = y{\bf{i}} + 3{x^2}{\bf{j}} - x{\bf{k}}$.
