Overview
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Do most of the solve using A-F rather than numbers, only resolve letters at the end.
I made r2c1 A and r2c4 F, with BCDE along the top between.

r1c[56] becomes an [AF] pair, rest of shape is [BCDE]
r[36]c2 becomes an [AB] pair, the rest of the shape is [CDEF].
However, r3c2 can't be A, as that leaves the middle region without an A.

We can conclude r3c1 is 2, but there's no benefit to marking it anywhere. Similarly, A and B are [14], and therefore monogamous with 2s on black kropkes.
Therefore r3c4 can't be A.
The bottom left region's, and therefore row 5's, A must be in r5c[36].
Therefore column 4's A can only be in r4c4.

middle pan region's B can only be in row 2, so top right region's B can only be in row 4, and middle elephant region can only be ib row 5
middle pan region's F can only be in r3c[35]
Therefore r3c[35] is an [AF] pair.
As r3c5 is [AF], it can only kropke connect to [CD] or [CDE], not B.
Which places row 4's B in r4c6.

If we hypothesise r3c5 is an F then it immediately can't pair with a C, as that removes all Cs from the elephant region.
However, if it pairs with a D, the r5c[23] kropke is forced to be DxF, and that forces both D and F into r6c1.
=> r3c5 is A, and you can complete the As.

A now pairs with [CDE], [CDE], and [DEF] on the 3 kropkes, so pairs with only [DE].
=> r4c2 is the elephant region's F
We can now associate (colour) r3c1, r5c2, and r4c5 together as the 2=[DE] that pairs with both 1 and 4 on black kropkes.
In the pan region, this [DE] can only go in r2c3, which is currently [BEC].
Therefore all four are E.
THerefore r6c6 is also E
=> you can letter the whole grid

We have AxExB from [124], FxC from [36], and FoB from [36]o[14]
=> F is 3, B is 4, and the rest follows.