Row 5 of this puzzle contains two highlighted cells, each of which contain "pencil marks" for 2 and 3 and
nothing else. This leads to the inescapable conclusion that one of the cells must contain a 2, and one must contain
a 3. We don't know at this stage which way it will be, but we do know that neither a 2 nor a 3 can appear elsewhere
in the 3x3 sub-square. Therefore, we can remove the "pencil mark" 3 from C7-R6.
This may not seem like a lot of progress, but it does uncover another valuable opportunity. C7-R6 and C7-R7 now
constitute another
Naked Pair which allows the 4 and the 6 to be removed from C7-R4. This leads to a
Single Candidate situation for C7-R4 which can immediately be solved.