How to solve this 6×6 logic puzzle

6×6 · severe · 17 steps

2🐱

🐱

This 6×6 board rates as severe (difficulty score 1118). Solving it from scratch takes 25 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 D5, 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 E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
After that, every remaining candidate in this region touches D3 and E3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Now, marking A1 would force E2 (the last open cell in region 2), then C3 (the last open cell in region 3), and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
From there, if B1 were the marker, it would force E2 (the last open cell in region 2), then C3 (the last open cell in region 3), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, that one can't be the marker: at C1 it would force E2 (the last open cell in region 2), and region 3 would be left with no open cell for its marker.
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, together, two regions have open cells only in rows 2 and 3; 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.
Then, if E1 were the marker, it would force D4 (the last open cell in region 5), then F3 (the last open cell in region 3), and region 6 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 E6 (the last open cell in region 6), then D4 (the last open cell in region 5), and region 3 would be left with no open cell for its marker.
Now, 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. Placing here also clears its 1 touching neighbour.
From there, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region. Placing here also clears its 2 touching neighbours.
Following that, place a marker at F6. Region has exactly one open cell left. Placing here clears the rest of row 6, column F, and its region.
First, place a marker at C3. Region has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region. Placing here also clears its 1 touching neighbour.
Next, place a marker at A2. Region has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
Finally, place a marker at B5. Region has exactly one open cell left.

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