A chain on one digit has the digit at one end or the other, so a cell that sees both ends cannot hold it.
Where to lookA pointer first: link the marked cells into a chain through 3.X-ChainFollow the chain: either row 3, column 9 is 3 or row 6, column 4 is 3, at least one of them. So no cell that shares a group with both ends can be 3.X-ChainFollow the chain: either row 5, column 5 is 1 or row 8, column 4 is 1, at least one of them. So no cell that shares a group with both ends can be 1.
Pick a digit and find regions where it has only two places. One of the two must be right. Join such pairs into a chain, linking them through cells that share a region, strong and weak links alternating. One end of the chain always holds the digit, so remove it from cells that see both ends.
This example, step by step
Where to lookA pointer first: link the marked cells into a chain through 3.
X-ChainFollow the chain: either row 3, column 9 is 3 or row 6, column 4 is 3, at least one of them. So no cell that shares a group with both ends can be 3.
X-ChainFollow the chain: either row 5, column 5 is 1 or row 8, column 4 is 1, at least one of them. So no cell that shares a group with both ends can be 1.
In hints
The hint first gives a direction. Blue marks the cells of the chain, and orange strikes remove the digit from the cells that see both ends.