The first clue tells you that all females besides two in the first column must be adjacent to exactly three other females. That clue alone will tell you almost everything you need to know in this puzzle.
Take a look at the right corners. If one of them is a female, then you’d have to have a 2×2 box of females so that the corner could be adjacent to three other females. Of course every female in that 2×2 box would then be adjacent to the other three. But one of them would be adjacent to the female in the middle column, thus adjacent to four females. So neither corner in the right column is female; both are male.
Any cell adjacent to eight cells cannot be adjacent only to females, because several of those females would then be adjacent to more than three females. In fact, the same is true of the cells that are adjacent to five cells (outer edge). The only cell that can be adjacent only to females, and still none of those females would be adjacent to more than three other females, is the final corner. So the bottom right corner is referring to the bottom left corner. It is a male, with all its adjacent cells being females.
The female in the bottom row is adjacent to two females. The third one must be one of the two cells to its left: C4 or C5. But either one of those will also be the third female adjacent to the female at B4. So those two females will certainly share a third adjacent female, meaning the first three cells of the middle row must all be males.
C4 or C5 is a female. If C4, then it would already be adjacent to three other females, meaning everything else adjacent to it would be male. If C5 is female, it would still need to be adjacent to one more female.Obviously it couldn’t be C4 or several females would be adjacent to four females. It couldn’t be D5 either, because D5 itself couldn’t be adjacent to three other females without making C5 adjacent to a fourth female. So whether C4 or C5 is the third female adjacent to the two females in the second column, either way D5 is male.
If E4 is female, then you’d have a 2×2 square of females. But the one at D4 would be adjacent to the other three in the square and either C4 or C5. So E4 cannot be female; it is male.
You still don’t know whether C4 or C5 is the third female adjacent to B4 and B5. If C4 is female, it is already adjacent to three females, making C5 and D4 male. D3 would still need two more female adjacents. E3 would not work because that would force a 2×2 female box, meaning D3 would be adjacent to more than three females. So E3 would be male. E2 couldn’t be female either because it would force D1 and D2 to be female. D2 would be adjacent to three females already, but D1 would not, and couldn’t without forcing D2 to be adjacent to a fourth. So if neither E2 nor E3 would work as females, both C2 and D2 would have to be females in order for C3 to be adjacent to three. Then C2 and D2 would be adjacent to two females each, and they’d clearly have to share the third one: either C1 or D1. But neither C1 nor D1 could be adjacent to three females without making C2 or D2 adjacent to more than three. So this entire scenario, which started with C4 as female, does not work. See below:
So C4 is not female, which means C5 is.
B4 and B5 are now adjacent to three females, so C4 is male.
C5 needs a third female adjacent, so D4 is female.
D4 needs a third female adjacent, so E3 is female.
E4 needs a third female adjacent: either D2 or E2. Both of those are also adjacent to D3, which also needs a third female adjacent. This means C2 cannot be female; it’s male.
If E2 is female, it would need one more female adjacent. That would have to be D1 since D2 being female would make D3 and D4 adjacent to four females. But if D1 were female, it could not be adjacent to three females without making D3 and D4 adjacent to four females either. So the third female adjacent to E4 is not E2; E2 is male. The third female adjacent to E4 must be D2.
D2 needs a third female adjacent: either C1 or D1. D1 cannot even be adjacent to three females, so it is male and C1 is female.
C1 needs two more female adjacents and is only adjacent to two unknown cells, so B1 and B2 are both female.
B1 and B2 are both already adjacent to three females, so A2, being adjacent to both of them, must certainly be male.