A Hard 6×6 Puzzle Solved with Forced Chain in 18 Steps

This 6×6 board rates as hard (difficulty score 1098). Solving it from scratch takes 18 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every remaining candidate in this region touches B2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, every remaining candidate in this region touches E2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches B4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, marking A1 would force B5 (the last open cell in region 4), then F2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then C3 and D3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), and the chain continues, and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, that one can't be the marker: at C1 it would force B3 (the last open cell in region 2), then F2 (the last open cell in region 3), then A5 (the last open cell in region 4), and region 6 would be left with no open cell for its marker.
6. From there, together, two regions have open cells only in columns B and A; 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.
7. Following that, together, two regions have open cells only in rows 1 and 2; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
8. 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.
9. Next, every remaining candidate in this region touches E4 and F4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
10. Then, every remaining candidate in this region touches E6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
11. After that, together, two regions have open cells only in columns D and C; 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.
12. Now, marking D1 would force A2 (the last open cell in region 1), then B5 (the last open cell in region 4), and then row 4 would be left with no open cell for its marker. So D1 can't be the marker there.
13. From there, place a marker at B1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at D2. Region has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at F3. Region has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
16. Next, place a marker at E5. Region has exactly one open cell left. Placing here clears the rest of row 5, column E, and its region.
17. Then, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
18. Finally, place a marker at C6. Region has exactly one open cell left.

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