Overview:
=========
- spot the trick
- top left
- row/col 1
- fill

Spot the Trick
--------------
This is equivalent to a Phistomefel ring, just rotated by 3 rows and columns:

...|A..|..A   Box
...|A..|..A    5   6   4
...|A..|..A
---+---+---
AAA|A..|..A
...|.BC|BC.    8   9   7
...|.CB|CB.
---+---+---
...|.BC|BC.
...|.CB|CB.    2   3   1
AAA|A..|..A

Therefore most of teh pattern of kropkes, the A cells in that diagram, are exactly teh same values, and thus modularly-restricted as the checkerboard pattern of residues B&C.

Immediately, you can't have 3x6 off the black kropkes, as both other residues crop up on them, and we can't have all three residues represented.
Just as immediately, you can't have 3x6 on the black kropkes, as r1c1 would need to be both 3 and 6.

=> The A, B, and C cells are all remainder 1 and remainder 2 modulo 3.

Pencil the black kropkes in A with [1248]
Remove 1 from all but two of them, the ones that also have a white kropke, as 2-1-2 is impossible.
Remove 1 from the others, as 1-2-3 is also forbidden.
All 4 possible 4s are accounted for.
On white kropkes, and with the no-[369] restriction [248]s pair with [157]s.
The remaining three A cells start as [12578]

Top Left
--------
The black kropke in box 1 can't have a 4, and can't be a 1x2 connecting to a 1, so is a 3x6 and can be placed.
1+2 can be removed from r4c[12], by the entropic line the same happens at the other end in column 4
In row 1, a 5 can't connect to +4+3 along the kropkes, nor +6+7, as both 3 and 6 are taken
=> r1c4 is 7 and connects inward +8+9, and downward +8x4+5+4x8+7.
Cleaning up the pencil markings in column 9 gives 5+4x[28]-[12]
Cleaning up row 9 gives 5+4x2+1
Entropy in r1c1 only permits a 4
Colour the known cells for modularity

Row/Column 1
------------
r7c1 on the modular line is [258],but not 2 or 5, so is 8.
r1c7 on the modular line is [258], but not 8 or 5, so is 2.
=> r3c9 is 8 and r9c9 is 7
Count the mod1s and the mod2s on the A cells - there are 8 mod1s and 7 mod2s
=> r4c9 is the remaining mod2, and therefore is 2

Therefore the B&C cells are 14444777 and 22555888 one way round or the other

The remaining [129] in column 1 can't have teh 2 in box 7, but in box 4 the 1 must go on the modular line
=> r8c1 is 1, and r[56]c1 is a [29] pair.

Similarly, the remaining [136] in row 1 can't have the 1 in box 2, as 1 must be on the modular line
=> r1c8 is 1, and r1c[56] is a [36] pair.
=> in box 2 the modular line is a [19] pair.

Fill
----
In row 3, there's a [1379] quad, so r3c2 isn't 1
On the entropic line, either r6c2 or r2c6 is 1, so r2c2 can't be 1.
Similarly, either r5c3 or r3c5 is 1, so r3c3 can't be 1.
=> box 1 completes, and pencil marking cleanup continues into box 3 and 4
=> r6c3 is 5

We know the B&C block contains two of each of the non-3 mod families in every row and column
=> you can remove 4&7 from r8c2, placing box 7's 7 in r7c2
=> you can remove 1&4 from r7c9, placing box 9's 1 in r7c7
=> r8c8 is the remaining [147], the 4.
=> The other 6 B cells are at best [47], but quickly resolve into 7s and 4s uniquely
The 8 C cells can be at best [258], but quickly resolve into 8s, 5s, and 2s uniquely
Everythign apart from teh [369]s resolves immediately.
The box 9 entropy line removes 6 from its ends, placing a 6 in r9c7, and you're done.

Proof of uniqueness
-------------------
https://sigh.github.io/Interactive-Sudoku-Solver/?q=.Pair%7EJRSkMJSikJRSkE%7E_isomodular%7ER8C5%7ER7C6%7ER6C5%7ER5C6%7ER6C7%7E%7ER5C8%7ER6C7%7ER7C8%7ER8C7%7ER7C6%7E%7ER5C5%7ER6C6%7ER5C7%7ER6C8%7ER7C7%7E%7ER8C8%7ER7C7%7ER8C6%7ER7C5%7ER6C6.BlackDot%7ER4C2%7ER4C3.BlackDot%7ER2C4%7ER3C4.BlackDot%7ER2C9%7ER3C9.BlackDot%7ER9C2%7ER9C3.BlackDot%7ER3C1%7ER2C1.WhiteDot%7ER4C2%7ER4C1.WhiteDot%7ER1C4%7ER2C4.WhiteDot%7ER1C9%7ER2C9.WhiteDot%7ER9C1%7ER9C2.WhiteDot%7ER4C3%7ER4C4.WhiteDot%7ER3C4%7ER4C4.WhiteDot%7ER4C1%7ER3C1.WhiteDot%7ER1C4%7ER1C3.WhiteDot%7ER1C3%7ER1C2.Entropic%7ER4C1%7ER3C2%7ER2C3%7ER1C4.Entropic%7ER1C1%7ER2C2%7ER3C3%7ER4C4.Entropic%7ER7C9%7ER8C8%7ER9C7.Modular%7E3%7ER7C1%7ER6C2%7ER5C3%7ER4C4%7ER3C5%7ER2C6%7ER1C7