Solution: Given that order of group $G$ is $9$. Since $o(G)=3^2$, so any non-identity element of $G$ must have order $3$ or $9$ (using the Lagrange's Theorem). If there is an element whose order is $9$, then $G$ will be cyclic and hence abelian. So, let us assume that order of every non-identity element is $3$. Let $x\in G$ and take $y\in G\setminus \langle x \rangle $. So, the elements of $G$ will be of the form $x^i y^j$ for $i,j \in \{0,1,2\}$. It is enough to show that $xy=yx$.

Let $yx=x^iy^j$ for some $i,j\in \{0,1,2\}$. If either of $i$ or $j$ is zero, then we will get $y\in \langle x \rangle $, a contradiction. So, $i$ and $j$ both are non-zero. We will show that all other cases are not possible. \begin{align*} yx = x^2 y & \implies yxy^{-1} = x^2 \\ & \implies y^2xy^{-2} = y\left( x^2 \right)y^{-1} = \left( yxy^{-1} \right)^2 = x^4 \\ & \implies y^{3}x y^{-3} = x^8 \\ & \implies y^3 x y^{-3} = x^2 \\ & \implies x = x^2, \end{align*} a contradiction.

Similarly, \begin{align*} yx = xy^2 & \implies x^{-1}yx = y^2 \\ & \implies x^{-3}yx^3 = y^8 = y^2 \\ & \implies y = y^2, \end{align*} a contradiction.

Finally, if \begin{align*} yx = x^2 y^2 & \implies yx = x^{-1} y^{-1} = (yx)^{-1}, \end{align*} which implies order of $yx$ is either $1$ or $2$, any of this is not possible ($1$ is not possible because it can not be identity and $2$ is not possible due to Lagrange's theorem). Therefore, none of the possibility is true and hence, $yx=xy$. Thus, we proved that $G$ is an abelian group.