Assume there is a box B which contains a set of cells C that make up a naked subset V with exactly one extra candidate x in one of the cells. The cell with the extra candidate lies in a row or column R, which also contains a cell Z with exactly two candidates; one in V, and one x. You can then eliminate the common candidate from B & R - C (the box-row intersection minus the cells C).
In less techinical terms, find a bivalue cell where setting one candidate directly eliminates another, and setting the other unlocks a naked pair/triple/quad that does the same elimination. (If it unlocks a naked pair, it's an XYZ-Wing.)
An example:
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3 5 4 | 68 689 689 | 2 7 1
6 9 8 | 17 17 2 | 3 4 5
27 27 1 | 5 3 4 | 8 6 9
----------------------+-----------------------+----------------------
89 27* 279*| 127 1278 3 | 4 5 6
58 1 3 | 68 4 568 | 7 9 2
4 6 257-| 9 27 57 | 1 8 3
----------------------+-----------------------+----------------------
27 8 6 | 27 5 1 | 9 3 4
1 3 79^| 4 679 679 | 5 2 8
59 4 259 | 3 289 89 | 6 1 7
This is just a typical XYZ-Wing:
C=R4C23, V=27, x=9, R=C3, Z=R8C3
-> Eliminate 7 from R6C3
Another one, slightly more interesting;
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1 4 2 | 8 69 356 | 35 39 7
9 7 3 | 1 24 245 | 58 48 6
5 8 6 | 34 49 7 | 2 349 1
----------------------+-----------------------+----------------------
7 3 8 | 2 5 9 | 1 6 4
4 6 1 | 7 3 8 | 9 2 5
2 9 5 | 46 1 46 | 7 38 38
----------------------+-----------------------+----------------------
6 5 47^| 9 478* 34-| 38 1 2
3 1 49 | 5 248* 24*| 6 7 89
8 2 79 | 36 67 1 | 4 5 39
The XYZ-Wing's bigger brother.
C=R7C5+R8C5+R8C6, V=248, x=7, R=R7, Z=R7C3
-> Eliminate 4 from R7C6
I wouldn't be surprised if someone thought of this before, nor if there is another technique that can deduce the same eliminations, but I figured I should give it a quick writeup anyway. Hope it can be useful to someone.
Now, off to find a quad+x!