Answer
\begin{align*}
\mathbf{div(F)} &=1-4xz-x+2zx^2\\
\text{curl} \mathbf{(F)}&=\left( -1\right)i - \left( 2xz^2-2x^2\right)j + \left( -y\right) k
\end{align*}
Work Step by Step
Given $$\mathbf{F}=\left\langle x-2 z x^{2}, z-x y, z^{2} x^{2}\right\rangle$$
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 ( x-2 z x^{2})}{\partial x}+\frac{\partial ( z-x y) }{\partial y}+\frac{\partial ( z^{2} x^{2})}{\partial z}\\
&= (1-4xz)+(-x)+(2zx^2)\\
&=1-4xz-x+2zx^2
\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}\\
x-2 z x^{2}& z-x y& z^{2} x^{2}
\end{array} \\
&= \left( \frac{\partial( z^{2} x^{2} )}{\partial y}- \frac{\partial( z-x y)}{\partial z}\right)i - \left( \frac{\partial( z^{2} x^{2} )}{\partial x}- \frac{\partial( x-2 z x^{2} )}{\partial z}\right)j + \left( \frac{\partial( z-x y )}{\partial x}- \frac{\partial( x-2 z x^{2} )}{\partial y}\right) k\\
&= \left( -1\right)i - \left( 2xz^2-2x^2\right)j + \left( -y\right) k\\
\end{align*}
