There are two ways to measure output impedance: Zout=dVout/dIout
Make the circuit hold its voltage constant, inject an AC current I into the output, and measure output voltage variation V. Then, Zout=V/I.
Input an AC voltage at the input, measure output voltage. Connect a known load, say a resistor that will draw a current from the output. Compute the difference dV in output voltage with and without output current. Zout=dV/dI.
These two are the two methods you're talking about in the question (although you have to squint a bit to recognize the second one in the scan from the book).
The second is much more difficult because you have to do two measurements and substract. If output impedance is small, the two measurements will be very close together, which introduces error. The first method is a more practical and direct measurement, so that's usually how it's done.
You can also do it with a DC current, but it is much easier with AC, as measuring a small AC voltage is easy and ignores any DC variations due to offset, temperature, low frequency noise, etc. Also, output impedance depends on frequency, so if you're interested in that, you have to measure it using AC and do a frequency sweep.
If you have good reason to believe the output impedance will be constant over a wide range of DC output voltage, then the actual value of the output voltage doesn't matter as long as it is constant. The simplest is to just use 0V, which means to connect the input of the opamp to ground.
This is usually the case for an opamp, as long as it's not clipping: output impedance won't change much with output voltage. It will vary quite a lot with DC output current, though, because the transconductance of output devices depends on current.