Overview
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This is a pure colouring puzzle, even though you use the numbers to decide what cells can be which colour.
The biggest "in" is recognising that the wide staples split into a set of four and a set of two.

Immediately, r1c1 is r3c2 and r2c1 is r6c2 else the middle pan region breaks. 
Ditto r[56]c6 and r[14]c2.
A similar relation exists for r6c[12] and r5c[36], and r1c[56] and r2c[14], and those are disambiguated not by the central regions, but by the '<' relations.
=> r2c1 windmills all the way around the grid, and by '<' it's [43], with [23] and [12] on its '<' side, and [45] and [56] on the '>' side.
By now you should have 5 colours, but no proof that r1c1 isn't the same as r6c6.

r1c6, being [12] colours one of r3c[35] in the middle pan region, and one of r6c[35] in the bottom region.
=> it x-wings r[45]c[12], and can't be r5c2 which is [45].
=> it's r5c1 and r4c2, colour and number them [12]
=> semi-colour the remaining x-wing

Similarly, r6c1, being [56] is pushed into r4c4, r3c5, r2c5, and r1c2. Colour and number them [56].
=> r3c3 and r6c5 are the same [12] as r1c6
=> r4c3 and r3c4 are the same [34] as r1c5

r2c5, being [23] and not the same colour as anything it can see, is pushed into r1c1, r3c2, r4c6, r5c4,and r6c3, so colour those the same as r1c1, and label them [23].

Similarly, r5c2, being [45] and not the same colour as anything it can see, i pushed into r1c4, r2c3, r3c1, r4c5, and r6c6, so colour those the same as r6c6 and label them [45].

6 cells remain uncoloured, and they are [45] by intersection of the sets they can be.
However, by r[45]c5, they are the 4 below a 5
Now just runaround  the 3, 2, and 1 below, and the 6 above