Suppose computing the 3D curl of a vector field involves finding this determinant:
$$ \mathrm{det} \begin{bmatrix} \mathbf{\hat{i}} & \frac{\partial}{\partial x} & 7y^3z^2 \\ \mathbf{\hat{j}} & \frac{\partial}{\partial y} & xyz \\ \mathbf{\hat{k}} & \frac{\partial}{\partial z} & -3x-z \end{bmatrix} $$
Using cofactor expansion, we should be able to break this into the subdeterminants obtained by excluding any row or column of our choice. However, if we choose the right column, the determinant will incorrectly appear as $(7y^3z^2 - xyz + (-3x - z))\mathbf{0} = \mathbf{0}$, the zero vector.
This is because the unit vectors have nothing but constant entries, thus their partial derivatives will always equal $\mathbf{0}$. Choosing the left column is the sane route, while choosing the middle column appears to work, but choosing the right column leads to a different result. How is this consistent with the typical rules of determinants?
If you're going to use this method, you have to be careful to preserve the order of the terms at all times, because the "multiplication" is not commutative in this context! You must have a unit vector, followed by a partial derivative, followed by one of the polynomials in the right column. The partial derivative applies only to the polynomial, not to the unit vector. The unit vector just comes along for the ride. In particular, the following is incorrect:
Let's say we exclude the right column, as you did. Then, for the $7x^2y^2$ term, the correct computation is:
$$j \frac{\partial}{\partial z} 7x^2y^2 - k \frac{\partial}{\partial y} 7x^2y^2$$ which equals $-k (14x^2y)$.
Now repeat for the other two factors in the third column.