Regarding positive/negative feedback, the answer is not so simple. Remember that each "negative" feedback will turn into positive feedback for rising frequencies due to unavoidable phase shift of the amplifier. More than that, positive feedback allows stable operation as long as the (positive) loop gain is below unity.
Therefore, based on the classical feedback model the following definition gives the distinction between positive and negative feedback:
(1) The denominator of the closed-loop function is D(s)=[(1-LG(s)] with LG(s)=loop gain
(2) Negative feedback for |1/D(s)|<1 and positive feedback for |1/D(s)>1|
Hence, pos. fedback will enhance the closed-loop gain if compared with the gain without feedback (and vice versa for negative feedback).
In the given example (active 2nd-order high-pass stage) we have a fixed negative feedback loop and a positive frequency-dependent feedback loop. However, the net feedback will always be negative. It is the task of the positive loop to decrease the overall net negative feedback within a certain frequency range (around the pole frequency of the C-R network).
The example shown (highpass stage) allows to define three different feedback loops (because we can define three different openings) and, hence, three different loop gain expressions.
1.) The above considerations apply to the case where the "naked" opamp is considered as active unit. Hence, both feedback loops are to be opened at the same time directly at the opamp output node.
In the two following cases , only one of the two remaining loops are to be opened.
2.) As an alternative, we can consider the opamp with the resistive negative feedback as a "gain-of-two" amplifier and open only the frequency-dependent feedback loop with the two R-C sections. A visual inspection reveals that the "gain-of-two" amplfier now has positive feedback only. This is confirmed by investigation (simulation) of the expression |1/D(s)| which is always larger than unity.
3.) As another alternative, we can define a frequency-dependent block (opamp together with the positive R-C feedback path). In this case, which is a theoretical case only, we open the resistive feedback path only and investigate (simulate) again the expression |1/D(s)|.
As a result, we will see that there is negative feedback - except a small frequency region around the pole frequency of the filter (app. 2 kHz) where the negative feedback turns into positive feedback. See the attached graph which shows the expression |1/D(s)| .