A Hard 6×6 Puzzle Solved with Forced Chain in 16 Steps

This 6×6 board rates as hard (difficulty score 1066). Solving it from scratch takes 16 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
2. Next, together, two regions have open cells only in columns A and B; with one marker each, those two markers have to fill exactly those two columns, so every other region's open cells there can be cleared.
3. Then, every remaining candidate in this region touches D2 and E2 and D4 and E4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches B5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, marking A1 would force F2 (the last open cell in region 3), then D3 (the last open cell in region 4), and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
6. From there, every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
7. Following that, that one can't be the marker: at C1 it would force A2 (the last open cell in row 2), then B6 (the last open cell in region 6), then D5 (the last open cell in region 2), and the chain continues, and row 4 would be left with no open cell for its marker.
8. First, every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
9. Next, if E1 were the marker, it would force D3 (the last open cell in region 4), then A2 (the last open cell in row 2), then B6 (the last open cell in region 6), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at F1 it would force D3 (the last open cell in column D), then A2 (the last open cell in row 2), and row 4 would be left with no open cell for its marker.
11. After that, place a marker at D1. Region has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region. Placing here also clears its 1 touching neighbour.
12. Now, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region. Placing here also clears its 1 touching neighbour.
13. From there, place a marker at A2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
14. Following that, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at C4. Region has exactly one open cell left.
16. Finally, place a marker at F5. Region has exactly one open cell left.

The key move here was Forced chain. Once you can spot that, this board falls quickly. ∎