Using nabla with partial derivatives and the Laplace operation $\partial_x^2+\partial_y^2+\partial_z^2$ Source of the problem p.812 here. Suppose 
$$\bar{F}(x,y,z)=(xy-z^2)\bar{i}+(xyz)\bar{j}+(x-y^2-z^2)\bar{k}.$$
I am concerned where I need to nabla an unit vector for example with 
$$\triangledown \times F.$$
I am particularly uncertain with $\partial_{x}\bar{k}$ (where $\bar{k}$ is an unit vector) -- is it zero or $\bar{k}\partial_{x}$?
This scan reuses the same logic and if the latter is wrong, it will have many errors (particularly in the middle section)

 A: 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.
A: The question is not whether $\bar k$ is a unit vector but whether it depends on $x$. From your scanned page, it seems that it's one of the canonical basis vectors and thus doesn't depend on $x$; thus its derivative with respect to $x$ vanishes.
The main problem in your calculation seems to be that you write $\nabla\times \overline F$ but then go on to calculate not $\nabla\times \overline F$ but $\nabla\times (\nabla(\nabla\cdot\overline F)).$
