Do you mean for $33$ or $121$ to be counted as prime numbers?
Note that the Sieve of Eratosthenes works by successively eliminating multiples of $2,3,5,7,11 \dots$
Once you have got to the prime $p$ you can be sure that any number left which is less than $p^2$ is prime.
Your list fails in two respects. First it is a finite list, but there are infinitely many primes (a fact known to Euclid). Take a prime not on your list, say $11$, and its square will not be prime.
Secondly you have chosen $9$, which is not prime. This doesn't make a huge difference here, because a number is either a power of $3$, in which case it is eliminated (except for $1,3$), or it has a prime factor other than $3$, in which case it would be eliminated by that prime (hence my example of $33$).
There would be a real problem, though, if you tried to work with squares of all the primes - say $9, 25$ when you would be counting $15$ as a prime.