How to determine if the feedback is positive or negative in this circuit?
I simply go, and look at the circuit intensely (yes, that is a very important engineering technique) and then ask myself:
When I &quot;magically&quot; increase the output of the amplifier, does the positive input increase, or the negative decrease? Then, it's positive feedback.
If the positive input decreases when you increase the output, or the negative input increases, then it's negative feedback.
Clearly, your upper 10 kΩ resistor (label your darn components!) forms a positive feedback loop.
Clearly, also, there's some negative feedback through your 2 kΩ resistor and your lower diode (label your darn components!).
Which one is dominant under which input conditions (and possibly, history, because I think this circuit has &quot;memory&quot;) isn't obvious to me, even when I look at this for longer. So, I think in this very peculiar circuit! (Where did you encounter this? It's pretty interesting, and it doesn't look like educational material – unis typically pay attention to not use K like Kelvin where k like kilo was meant.)
Anyway, simulation would probably be the way to go here, because the diode forward voltages might play a big role in practice, and things get nonlinear. I don't think you can just work with the circuit as is without specifying the diode types, too.

One way to approach such atypically complex circuit would be to hypothesize about it, and test the hypothesis. You'll have to test all hypotheses you make afterwards for being true, and all you can then say is that you found one configuration for which some statement (positive or negative feedback) is true. You haven't proven anything in general.
Let's do that:

I proclaim there's a Vin at which no current flows through your 5 kΩ resistor (label your darn components!).
If no current flows through that 5 kΩ resistor (label your darn components!), then Vin = V- of the opamp.
I'll also assume -12V &lt; Vo &lt; +12V for a start. (there's nothing in the circuit specifying that. It could not be the case at the moment we switch our input voltage to the specific Vin mentioned above – and then, everything works differently. Which is why I said this circuit has memory.)
It's therefore practical to assume that Vin &lt; the voltage at the junction between your 3 kΩ resistor (label your darn components!) and your 1 kΩ (label your darn components!). So, the top diode (label your darn components!) is in reverse bias and doesn't conduct any electricity.
The same applies to the bottom diode (you know what I'm saying in these parentheses)
In that point, only positive feedback exists. Let's figure out what Vo is for that point, and whether our assumptions are right.
In that point of operation the current flowing through the 3 kΩ resistor is the same as through the 1 kΩ resistor (...) and the current through the 2 kΩ the same as through the 7 kΩ resistor. The former is Itop= (12 V - Vo) / 4kΩ, the lower Ibottom=(Vo + 12 V) / 9 kΩ.
The current Io flowing through the 10 kΩ resistors hence must be Itop-Ibottom; current can go nowhere else
But
\begin{align}
V_o &amp;= (2·10\,\text{kΩ})·I_o\\
&amp;= 20\,\text{kΩ} · (I_{top} - I_{bottom})\\
&amp;= 20\,\text{kΩ} · \left(\frac{12\,\text{V} - V_o}{4\,\text{kΩ}} - \frac{12\,\text{V} + V_o}{9\,\text{kΩ}}\right)\\
&amp;= 20·\left(\frac{12\,\text{V} - V_o}{4} - \frac{12\,\text{V} + V_o}{9}\right)\\
&amp;= \frac{240\,\text{V} - 20V_o}{4} - \frac{240\,\text{V} + 20V_o}{9}\\
&amp;=60\,\text{V}- 5 V_o -\frac{240}9\,\text{V} +\frac{20}9 V_o\\
-\frac{16}9 V_o &amp;= \frac{100}3\,\text{V}\\
 V_o &amp;= -\frac9{16}\frac{100}3\,\text{V}\\
&amp;=-\frac{300}{16}\,\text{V}\\
&amp;=-18.75\,\text{V}
\end{align}
which breaks our assumption above that the lower diode is reverse biased.

So, even that &quot;logical hypothesis&quot;-based testing fails, too. And that's all I'm willing to say here: there's no obvious working point where it's clearly one, or the other feedback.