Initial step in proving that a partial derivative of a tensor is not a tensor Exercise 91 in Pavel Grinfeld's book Introduction to Tensor Analysis and the Calculus of Moving Surfaces reads

Similarly, for a contravariant tensor $T^i,$ derive the transformation rule for $∂ T^i ∕∂ Z^j,$ and show that it is not a tensor.

The solution (here) seems to proceed by contradiction from
$$S^{ij}\overset{?}{=}∂ T^i ∕∂ Z^j$$
I don't understand this. The contravariant tensor with one single live index is really a vector (and $T^i$ would be the components of the vector), and the operation on the RHS is the partial derivative of its components with respect to the different coordinates.
This is not a tensor, because it depends on the specific coordinates, which may vary from point to point, and this change in basis vectors is not accounted for. There are a number of posts on this (example).
The question is - under the assumption that this was actually true, as in Euclidean space - why would it render a twice contravariant tensor, i.e. $S^{ij},$ as opposed to simply another contravariant tensor of order one, $S^i$?

As an aside, the notation in the book is confusing. He seems to use $Z_i$ or $Z^i$ for covariant and contravariant components, and $\bf Z_i$ and $\bf Z^i$ for covariant and contravariant basis vectors, but I believe the usual notation for basis vectors is $\partial_i$ (alternatively $\frac{\partial}{\partial_i}$) or simply ${\bf e}_i$) for covariant and $\rm dx^i$ (or ${\bf e}^i$) for contravariant. Is this right? The idea of the "Z" I guess is to get rid of any distraction from the indices...
 A: The gradient of a vector field is a tensor.
In Cartesian coordinates where there is no change in basis vectors from point to point this is clear:
$$\begin{align}
\nabla \vec v &= \frac{\partial}{\partial Z^1}\vec v\; \vec e_1 + \frac{\partial}{\partial Z^2}\vec v\; \vec e_2 + \frac{\partial}{\partial Z^3}\vec v\; \vec e_3 \\
&= \frac{\partial}{\partial Z^1}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_1 +\\
&+ \frac{\partial}{\partial Z^2}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_2 +\\
&+ \frac{\partial}{\partial Z^3}\left(v_1 \; \vec e_1 + v_2 \; \vec e_2 + v_3 \; \vec e_3 \right)\; \vec e_3 \\
&= \frac{\partial v_1}{\partial Z^1} \vec e_1\vec e_1 + \frac{\partial v_2}{\partial Z^1} \vec e_1\vec e_2 +\frac{\partial v_3}{\partial Z^1} \vec e_1\vec e_3 +\\
&+\frac{\partial v_1}{\partial Z^2} \vec e_2\vec e_1 + \frac{\partial v_2}{\partial Z^2} \vec e_2\vec e_2 +\frac{\partial v_3}{\partial Z^3} \vec e_2\vec e_3 +\\
&+\frac{\partial v_1}{\partial Z^3} \vec e_3\vec e_1 + \frac{\partial v_2}{\partial Z^3} \vec e_3\vec e_2 +\frac{\partial v_3}{\partial Z^3} \vec e_3\vec e_3 \\
&=\frac{\partial v_k}{\partial Z^j} \vec e_j\vec e_k\\
&=\frac{\partial v_k}{\partial Z^j} \vec e_j\otimes\vec e_k\\
&=T^{jk}
\end{align}$$
Therefore it makes sense to start off the prove that, besides in Cartesian coordinates, the partial derivatives of a tenson $T^i$ are not a tensor by assuming they do, and seeking a contradiction.
Found here.
