I am trying to generalise the question discussed by me in the following link and answered by lfz
Finite field with $9$th primitive root of unity. A $9$th primitive root of unity is an element of order $9$.
In a cyclic group of order $m$, there is an element of order $d$ iff $d$ divides $m$.
Therefore, as you have found, $GF(p^n)^\times$ has an element of order $9$ iff $p^n\equiv 1\bmod 9$.
Since $p$ cannot be $3$, we have $\gcd(p,9)=1$ and so $p^6\equiv 1 \bmod 9$ and $1 \equiv p^n\equiv p^{n \bmod 6}\bmod 9$.
If we are interested in which $n$ work for which $p$, then:
$n \equiv 1 \bmod 6$ implies $p \equiv 1 \bmod 9$
$n \equiv 2 \bmod 6$ implies $p^2 \equiv 1 \bmod 9$, that is, $p \equiv \pm1 \bmod 18$
$n \equiv 3 \bmod 6$ implies $p^3 \equiv 1 \bmod 9$, that is, $p \equiv 1 \bmod 6$
$n \equiv 6 \bmod 6$ implies $p^6 \equiv 1 \bmod 9$, that is, $p \equiv \pm1 \bmod 3$
The cases $n \equiv 4,5 \bmod 6$ reduce to $n \equiv 2,1 \bmod 6$ because $p^6\equiv 1\equiv p^n\bmod 9$ implies $p^{\gcd(n,6)} \equiv 1\bmod 9$.
Now QUESTION is can i say that All finite field $GF(p^m)$ has primitive $9$th root of unity where $m=6n+k$ and $n\ne 0?$
Discussion from the following link
The finite fields with a primitive $9$th root of unity are exactly the fields such that $\Phi_9(x)=x^6+x^3+1$ factors in linear terms. Assuming $p\neq 3$, the degree of the splitting field of $\Phi_9(x)$ over $\mathbb{F}_{p}$ is the least $h\geq 0$ such that $9\mid\left(p^h-1\right)$. Since $\varphi(9)=6$, any field of the form $\mathbb{F}_{p^{6k}}$ with $p\neq 3$ contains a primitive $9$th root of unity.