Here's another attempt to show a logical method of finding forcing chains rather than just giving them. I found this solution with pencil and paper. From the given position I was able to remove only a couple more candidates by typical means before locking for a forcing chain:

1) Ignore all cells except those with exactly two candidates.

2) Connect cells within BINARY GROUPS -- pairs, trips, quads, etc. with BLUE LINES. I define a "binary group" as a set of cells that when considered together, can exist in exacty two states. For example, three cells in a row with the candidates [12][23][13] can only exist as 1-2-3 or 2-3-1.

The binary groups are:

Rows: r1c1237, r3c1459

Columns: r138c1, r18c2, r235c4

Boxes: (r1c234-r2c3-r3c1), (r1c7-r3c9)

(Note: Lines between r1c1-r1c7, r3c1-r3c9, etc are assumed but omitted for clarity.)

3) Exactly one of r1c1 and r1c2 must be a '1'.

Exactly one of r8c1 and r8c2 must be a '1'.

This forms an x-wing in 1's. Though it won't allow us to exclude any candidates, it does allow us to connect r8c1-r8c2 with a blue line, as the four cells taken together have exactly two states -- 1-8-9-1 or 5-1-1-8.

4) Similar logic allows a blue line to be drawn between r2c3-r2c4. No more blue lines can be drawn.

This is one large binary group. A chain can be in any fashion folloing the blue lines without forming a proof. The entire structure can exist in two states without internal contradiction.

5) RED lines are drawn between all other possible connections. r8c1-r7c3-r2c3, r1c7-r8c7 and r3c9-r9c9.

To form a proof, either from forcing chains or proof by contradiction, a LOOP must be formed. Two cells must be connected by two paths, at least ONE of which must contain RED lines. A loop that follows only blue lines is a binary group and can exist in two non-contradictory states.

6) Though a loop has been formed that includes a red line through r7c3, it cannot be used because the candidate 6 doesn't pair with any other.

At this point, we come to the end of the logic I can do using my standard method for finding forcing chains using 2-candidate cells. No chain exists. One might say that it didn't work. I would argue that it did! It proved that a 2-candidate forcing chain *does not exist*, and I can stop looking for it and try something else. The limitation in this method is that it does miss other types of forcing chains that use cells with more than 2 candidates.

I'm forced to consider the cells with 3 candidates. r8c5 seems promising as each of its candidates match-up with a 2-candidate cell in its row or column.

7) r3c5 and r1c7 must BOTH be 3 or NEITHER is 3. This is because they are part of a binary group and are connected directly by r3c9.

a) If they are BOTH 3, then r8c7=6, then r8c6<>6.

b) If NEITHER are 3, then r3c5=6, then r8c6<>6.

I now eliminate candidate 6 from r8c6 and add RED lines between it and r8c2, r3c5 and r8c7.

The RED path through r8c7 cannot be used as the candidate 6 doesn't pair with any other.

This leaves one red path between r8c2 and r3c5.

8)

r8c2=8 => r8c5=3 => r3c5=6

r8c2=1 => r1c2=8 => r1c3=3 => r1c7=5 => r3c9=3 => r3c5=6

Therefore, r3c5=6

The rest is easy.