How does this MOSFET-configured chaos circuit work?
The left part of the circuit:

is an LC parallel circuit. If you would charge C2 to a certain voltage, then disconnect the voltage source and measure at VC2 you would see a damped sinewave. The sinewave will die out due to the loss introduced by R2.

If we could compensate for the losses introduced by R2 (and other losses as well) then we would have an oscillator and then the sinewave would not die out.
The way to do that is by adding a resistor with a negative value in parallel to the LC parallel circuit. The right part of the circuit with the MOSFETs and the battery behaves somewhat like that negative resistor. At least for small signals it does.
To prove that this circuit actually behaves as a negative resistor for small signals you would need to analyze it using small signal analysis.
Note that the battery needs to be there to provide power to the MOSFETs so that they can be biased and also to make the circuit able to deliver power to the other circuit.
An (ideal) negative resistor does not dissipate power (like a normal, positive resistor does) instead it delivers power!

The negative resistor isn't fully in parallel with the LC parallel circuit, instead R1 and C3 are used to &quot;decouple&quot; the circuit a bit from eachother. That way the LC circuit is influenced less by the behavior of the negative resistor part of the circuit.
The negative resistor only works properly for small signals, for larger signals the negative resistor's value will increase making it have less influence on the LC circuit. This is actually good because this will make the circuit oscillate at a certain amplitude that remains stable.
To learn more about LC oscillators, maybe this will help.