Remove duplicates from list and order it by count? (python 3)
44 bytes

sorted({}.fromkeys(s),key=s.count,reverse=1)

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An improvement to the below, replacing the key with a built-in rather than a lambda, and using reverse to swap the comparisons. We'd like to do [::-1] on the final result instead, but that doesn't do the right stable tiebreaks.

47 bytes

sorted({}.fromkeys(s),key=lambda y:-s.count(y))

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A simple transposition of your code, moving your {}.fromkeys de-duplication trick before the sorting. Since sorting converts to a list, extracting the keys from a dictionary, this saves the list() call of the original.
For ease of reference, the original code is:
53 bytes

list({}.fromkeys(sorted(s,key=lambda y:-s.count(y))))

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