I considered a 3-dimensional vector field, with Cartesian components, $u_i$. My aim was to show that $\partial u_i / \partial x_j$ is an order 2 tensor.
I start with showing that $\frac{\partial}{\partial x_i}$ is a vector. (Skip the next section if we accept this as granted.)
Given that $x_i=L_{ji}x'_j$: $$\frac{\partial x_j}{\partial x'_i}=\frac{\partial(L_{kj}x'_k)}{\partial x'_i}=L_{kj}\frac{\partial x'_k}{\partial x'_i}=L_{kj}\delta _{ki}=L_{ij}$$ Applying chain rule: $$\frac{\partial}{\partial x'_i}=\frac{\partial x_j}{\partial x'_i}\frac{\partial}{\partial x_j}=L_{ij}\frac{\partial }{\partial x_j}$$ This is the transformation property which to be satisfied by a tensor, so we find that $\frac{\partial}{\partial x_j}$ is a tensor.
So we this knowledge I do:
$$\Big( \frac{\partial u_i}{\partial x_j}\Big)'=\Big(\frac{\partial}{\partial x_j} \Big)'(u_i)'=L_{jq}\frac{\partial}{\partial x_q}L_{ip}u_p=L_{jq}L_{ip}\frac{\partial u_p}{\partial x_q}$$
So $\frac{\partial u_i}{\partial x_j}$ transforms as a tensor, so I conclude that $\frac{\partial u_i}{\partial x_j}$ is a tensor.
However if I do:
$$\Big( \frac{\partial u_i}{\partial x_j} \Big)' = \frac{\partial (u_i)'}{\partial (x_j)'}=\frac{\partial L_{ip}u_p}{\partial L_{jq}x_q}=\frac{L_{ip}}{L_{jq}} \frac{\partial u_p}{\partial x_q}$$
Multiply both sides by $L_{jq}$, then $L_{qj}$:
$$\Big( \frac{\partial u_i}{\partial x_j} \Big)' = L_{qj}L_{ip}\frac{\partial u_p}{\partial x_q}$$
This is not the same as our previous result (which was: $\Big( \frac{\partial u_i}{\partial x_j} \Big)' = L_{jq}L_{ip}\frac{\partial u_p}{\partial x_q}$). So using this second method, $\Big( \frac{\partial u_i}{\partial x_j} \Big)$ is not a tensor.
But I know it is.
My questions are:
1) Is my first attempt to show that $\frac{u_i}{x_j}$ is a tensor right?
2) Why did the second attempt yielded a result which is so-similar - but wrong - result? What can I learn from the second result? (Ie what step is wrong and why?)