How to solve this 6×6 logic puzzle

6×6 · severe · 19 steps

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This 6×6 board rates as severe (difficulty score 1139). Solving it from scratch takes 25 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every open cell left in this region sits in column A, so its marker has to land there; that clears every other open cell in the column.
Next, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
Then, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
After that, 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.
Now, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, every remaining candidate in this region touches B5 and D5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Following that, marking A1 would force F2 (the last open cell in region 2), then D3 (the last open cell in region 4), then E6 (the last open cell in region 6), and the chain continues, and then row 5 would be left with no open cell for its marker. So A1 can't be the marker there.
First, 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.
Next, marking D1 would force A2 (the last open cell in row 2), then C5 (the last open cell in row 5), and then column B would be left with no open cell for its marker. So D1 can't be the marker there.
Then, if E1 were the marker, it would force A2 (the last open cell in row 2), then C5 (the last open cell in row 5), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, place a marker at F1. Region has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
Now, that one can't be the marker: at A2 it would force C5 (the last open cell in row 5), and column B would be left with no open cell for its marker.
From there, place a marker at D2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
Following that, 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.
First, if A3 were the marker, it would force B6 (the last open cell in region 3), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, place a marker at B3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region. Placing here also clears its 2 touching neighbours.
Then, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
After that, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region.
Finally, place a marker at E4. Region has exactly one open cell left.

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