In line three, you should have $x-2$ in the third component rather than $-2$.
In line four, you computed the curl of your answer in line three (which is $\rm \nabla(\nabla\cdot F)$); instead you should have been computing the curl of the original function $\rm F$, as in
$\nabla\times \mathrm{F}=\begin{vmatrix} \color{Red}{\mathbf{i}} & \color{Red}{\mathbf{j}} & \color{Red}{\mathbf{k}} \\ \color{Blue}{\partial_x} & \color{Blue}{\partial_y} & \color{Blue}{\partial_z} \\ \color{Green}{xy-z^2} & \color{Green}{xyz} & \color{Green}{x-y^2-z^2} \end{vmatrix}$
$=\big(\color{Blue}{\partial_y}(\color{Green}{x-y^2-z^2})-\color{Blue}{\partial_z}(\color{Green}{xyz})\big)\color{Red}{\mathbf{i}}-\big(\color{Blue}{\partial_x}(\color{Green}{x-y^2-z^2})-\color{Blue}{\partial_z}(\color{Green}{xy-z^2})\big)\color{Red}{\mathbf{j}}+\big(\color{Blue}{\partial_x}(\color{Green}{xyz})-\color{Blue}{\partial_y}(\color{Green}{xy-z^2})\big)\color{Red}{\mathbf{k}}$
$=-(x+2)y\mathbf{i}-(1+2z)\mathbf{j}+(yz-x)\mathbf{k}.$
As a general note, since $\partial_x$ denotes partial differentiation with respect to $x$, and $\mathbf k$ is a constant vector, if we encountered $\partial_x\mathbf{k}$ it would be $(\frac{\partial}{\partial x}1)\mathbf{k}=\mathbf{0}.$
The Rule of Sarrus is invalid when you are mixing operators and vectors and scalar components together in a matrix. For each term in the expansion, you must put the differentiation operator to the left of the component so that it can act on it. Above this is depicted as blue-left-of-green.