How to solve this 6×6 logic puzzle

6×6 · severe · 21 steps

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This 6×6 board rates as severe (difficulty score 1211). Solving it from scratch takes 28 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.

First, 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.
Next, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, marking A1 would force E3 (the last open cell in column E), and then column D would be left with no open cell for its marker. So A1 can't be the marker there.
After that, if B1 were the marker, it would force D2 (the last open cell in region 2), and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, that one can't be the marker: place it at C1, and region 2 would be left with no open cell for its marker.
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.
Following 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.
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.
Next, every remaining candidate in this region touches B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, every remaining candidate in this region touches D4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
After that, if E1 were the marker, it would force A2 (the last open cell in row 2), then B6 (the last open cell in column B), then C4 (the last open cell in region 5), and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, that one can't be the marker: at F1 it would force A2 (the last open cell in row 2), then B6 (the last open cell in column B), then E5 (the last open cell in region 4), and the chain continues, and row 3 would be left with no open cell for its marker.
From there, 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.
Following that, every open cell in column F 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.
First, together, two regions have open cells only in columns C and E; 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.
Next, if F2 were the marker, it would force C3 (the last open cell in region 3), then A4 (the last open cell in region 1), and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, 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.
After 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.
Now, place a marker at F5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
Finally, place a marker at E3. Region has exactly one open cell left.

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