Hint:
Ignore the numeric implications of the renbans and Vs until the end, you can do this with letters quite easily.

Walk-through:
Label the top left region A below BCDE above F.
By sudoku, the top right region becomes AF above BCDE.
To not break the middle pan region, the bottom left region becomes B right of CDEF left of A.
r3c[12] equals r4c[56], so r3c1 can't be F, so is E.
r6c[34] is a [BE]v[CD] V, so r2c[56]'s [BCDE]v[CD] must also be [BE]v[CD].
=> The row 2 [BE] pair makes r2c2 a D, r4c2 an F and r5c2 an E
=> r2c6 is C, and because r2c5 and r6c3 are the same [BE] (both different from r2c3), r6c4 is also C
=> r5c4 is column 4's B
=> r3c4 can't be D in the region, so is A, and r4c4 is D
=> r5c5 is C, r4c3 is A and r4c1 is C
=> r5c3 is not C so is F, resolving all the [AF] and [DF] pairs in teh grid

At this point, only 6 [BE] cells exist.

ADB is a renban, and AC[BE] also.
If the upper AB[BE] renban was ACB, then [AB] are adjacent, and [DC] have difference 3 sandwiching [AB].
But that makes r4c6 also a B, breaking sudoku.
=> r2c3 is E

ADB is a renban, and ACE is a renban also.
Therefore A is 3 or 4, and B and C are opposite sides of it.
However B+C is 5, so A can't be 4, so is 3.
[BC] become a [14] pair.
CD3B is [14]D3[14] is a renban, so D is 2
E[14]3 is a renban, but E can't be 2, so C can't be 1, so is 4, and B is 1, and E can only be 5.
=> F is 6.

Bosh!