A Hard 6×6 Puzzle Solved with Forced Chain in 12 Steps

This 6×6 board rates as hard (difficulty score 1040). Solving it from scratch takes 12 logical steps, using 5 techniques: Single cell, Row/column exclusion, Adjacency clear, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every remaining candidate in this region touches D2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force B3 and B4 cleared (every open cell left in this region sits in column B, so its marker has to land there), then C2 and C3 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D3 (the last open cell in region 4), 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.
3. Then, if B1 were the marker, it would force E3 and F3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then C3 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D3 (the last open cell in region 4), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
4. After that, marking D1 would force C3 (the last open cell in region 4), and then region 3 would be left with no open cell for its marker. So D1 can't be the marker there.
5. Now, if E1 were the marker, it would force A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then B2, B3 and B4 cleared (every open cell left in this region sits in column B, so its marker has to land there), then C2 and C3 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, that one can't be the marker: at F1 it would force A5 and A6 cleared (every open cell left in this region sits in column A, so its marker has to land there), then B2, B3 and B4 cleared (every open cell left in this region sits in column B, so its marker has to land there), then C2 and C3 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and row 4 would be left with no open cell for its marker.
7. Following that, place a marker at C1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region. Placing here also clears its 1 touching neighbour.
8. First, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region. Placing here also clears its 2 touching neighbours.
9. Next, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
10. Then, place a marker at B4. Region has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region. Placing here also clears its 1 touching neighbour.
11. After that, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
12. Finally, place a marker at E5. Region has exactly one open cell left.

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