...54..32....629.5.25.9346.5.138924..4.276.5.2..415.9..3.92.5....2.513.4.58.3..29
I thought this puzzle was interesting. Having done no good in skyscrapers or kites I looked at the forced moves caused by the two possible locations of 3 in Box6. This led to the interesting result that in both cases there would be a locked 67 pair in Boxes 4 & 6, and these cells were all in Rows 4 & 6, forming a 4-cell bivalue chain, and in both outcomes three of the four cells were the same. The resultant two versions of the middle band look like this:
In the upper version the 8 could only go in r6c9 as putting it in r6c7 would cause a Unique Rectangle. Can this 4-cell bivalue loop be used to solve the puzzle, with 3 of its points known?