Answer
\begin{align*}
\mathbf{div(F)}&=0 \\
\text{curl} \mathbf{(F)}&=\left( 1-3z^2 \right)i - \left( 2x-1 \right)j + \left(1+2y \right) k
\end{align*}
Work Step by Step
Given $$\mathbf{F}=\left\langle z-y^{2}, x+z^{3}, y+x^{2}\right\rangle$$
Since
\begin{align*}
\mathbf{div(F)}&=\frac{\partial \mathbf{F}_x}{\partial x}+\frac{\partial \mathbf{F}_y }{\partial y}+\frac{\partial \mathbf{F}_z}{\partial z}\\
&=\frac{\partial ( z-y^{2})}{\partial x}+\frac{\partial (x+z^{3}) }{\partial y}+\frac{\partial ( y+x^{2})}{\partial z}\\
&= 0
\end{align*}
and
\begin{align*}
\text{curl} \mathbf{(F)}&=\begin{array}{|||}
i&j &k\\
\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
z-y^{2}& x+z^{3}& y+x^{2}
\end{array} \\
&= \left( \frac{\partial( y+x^{2})}{\partial y}- \frac{\partial( x+z^{3})}{\partial z}\right)i - \left( \frac{\partial( y+x^{2} )}{\partial x}- \frac{\partial( z-y^{2})}{\partial z}\right)j + \left( \frac{\partial( x+z^{3} )}{\partial x}- \frac{\partial( z-y^{2} )}{\partial y}\right) k\\
&= \left( 1-3z^2 \right)i - \left( 2x-1 \right)j + \left(1+2y \right) k\\
\end{align*}
