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 1208). Solving it from scratch takes 27 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 open cell left in this region sits in column F, 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 row 6, so its marker has to land there; that clears every other open cell in the row.
Then, together, two regions have open cells only in columns E and D; 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.
After that, 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.
Now, 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.
From there, 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.
Following that, marking A1 would force B3 (the last open cell in region 4), then C5 (the last open cell in region 5), and then column D would be left with no open cell for its marker. So A1 can't be the marker there.
First, every open cell in column A 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.
Next, that one can't be the marker: at C1 it would force B4 (the last open cell in region 5), and row 5 would be left with no open cell for its marker.
Then, if E1 were the marker, it would force D6 (the last open cell in region 6), then A5 (the last open cell in row 5), and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, that one can't be the marker: at F1 it would force B4 (the last open cell in column B), and row 5 would be left with no open cell for its marker.
Now, 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.
From there, every remaining candidate in this region touches E3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, 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.
First, marking E2 would force F4 (the last open cell in region 3), then C5 (the last open cell in region 5), and then row 6 would be left with no open cell for its marker. So E2 can't be the marker there.
Next, place a marker at F2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
Then, if A3 were the marker, it would force E4 (the last open cell in region 2), then C5 (the last open cell in region 5), and row 6 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 D3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region. Placing here also clears its 1 touching neighbour.
Now, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region.
From there, place a marker at A4. Region has exactly one open cell left.
Finally, place a marker at E6. Region has exactly one open cell left.

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