RMS Current through output diode in a flyback transformer with multiple outputs
I don't know where this formula comes from and I have never seen it. I derived the formula for the secondary-side diode rms current in my book published in 2014, page 755. I have built a quick Mathcad sheet to test it against your formula. This is a 72-W ac-dc adapter, no leakage inductance (otherwise the peak sec. current would be reduced) and low conduction losses in the MOSFET:

So for a 120-V input voltage, the sheet computes a peak current of 2.15 A and a valley current of 0.95 A in the primary. The rms current in the diode is computed to slightly less than 5 A.
To test these results, I have captured a quick flyback circuit with SIMPLIS and ran a simulation:

Once the simulation is done, pressing F3 in the curve windows gives you the diode rms current:

The simulated current is 4.98 A, very close to what the calculation gives. If I now use the formula you mentioned:

As you can see, there is no significant error with the simulation and my Mathcad sheet. So your formula is correct and elegantly arranged : )
One way to get the diode rms current is to determine the ripple in the magnetizing inductance of the flyback converter. You can follow the calculation details in a series of articles I published in How2Power in 2010. For a multi-output converter, precisely calculating all rms current can be somewhat adventurous considering all the leakage inductances at play which heavily distort waveforms (see here for the cantilever transformer model and Fig. 9 for the sec. waveforms).