Stream of consciousness here, might be hard to follow this post despite the simplicity of its its concepts due to lack of good formatting on my part. You have been warned
So... As to whether there can be any 9s...
Any solution with a 9 would have to have the 9 on one of the sides (truism).
The only possible 3 digit combinations for the 9 that could appear on one of the halves of the solution would have to be some combination from the set of 239 or 249. This is because 1 and 0 cannot be used and any other 3 digit set would violate rule 6 or 10.
2 cannot be adjacent to 3 (rule 7) or 4 (rule 5) so in both these scenarios 9 would have to be the second number of the 3 digit set to seperate the 2 from the other number.
This leaves us with:
293
392
294
492
9 and 3 cannot be adjacent though (rule 5) so we are left with:
These patterns must appear in any proposed solution with a 9. Can these patterns be ruled out?
In both cases the border of the other 3 digit set is an even number (2 or 4) and an even number cannot be adjacent to another even number (rule 8) so the next digit has to be odd. This leaves us with only 5 or 7 as the candidates for the first digit of the next 3 digit set (because rule 6 and 7 and excluding the number 1)
So now we know any solution containing a 9 must have this pattern of 4 digits:
2945
4927
2947
4925
But rule 7 once again comes in to spoil the fun, leaving us with:
Now let us look at remaining possibilities for the last 2 digits. Looking just at rule 6 we see that what we have to work with is this:
3 (5) 6 (7) 8
Parenthesis indicate numbers available in some of our partly-constructed solutions but not all.
Looking over these numbers we see 8 is excluded from filling out all our partial solutions due to rule 10. So...
3 (5) 6 (7)
All but one of these are odd numbers so we must be careful now of rule 9. In fact because of rule 9 the 6 MUST be used as one of the 2 final digits. The remaining possibilities are thus:
4927_6
2947_6
4925_6
49276_
29476_
49256_
All other iterations have been ruled out unless I made some mistake and the digits permitted to fill these blanks are:
3 (5) (7)
Except 3 and 6 cannot be adjacent (rule 8) so actually we can only fill each of these blanks with a 5 or 7, depending on which does not violate rule 6 for each possibility.
So now...
492756
294756
492576
492765
294765
492567
All of these violate rule 10.
Therefore unless I screwed something up 9 cannot exist in any possible solution.