Solution: Let the order of the group $G$ be $n$. According to the given hypothesis, $3$ does not divide $n$. Since, $3$ is a prime number, so we have \[ \gcd(3,n) = 1. \] The above implies, there exists $x,y\in \mathbb{Z} $ such that \begin{equation}\label{eq:01Mar2023-1} 3x+ny = 1 . \end{equation}

Now note that for any $a,b\in G$, \begin{align*} ab & = (ab)^{3x+ny} \\ & = (ab)^{3x} (ab)^{ny}\\ & = \left( (ab)^3 \right)^x \left( (ab)^n \right)^y \\ & = \left( a^3 b^3 \right)^x e^y \quad \textcolor{magenta}{\boxed{\text{as $|G|=n$}}}\\ & = \left( a^3 b^3 \right)^x \\ & = a^3 \underbrace{\left( b^3 a^3 \right) \left( b^3 a^3 \right) \cdots \left( b^3 a^3 \right)}_{(x-1)-\text{times}} b^3 \\ & = a^3 \left( b^3 a^3 \right)^{x-1} b^3 \\ & = a^3 \left( ba \right)^{3x-3} b^3 \quad \textcolor{magenta}{\boxed{b^3 a^3 = (ba)^3}} \\ & = a^3 (ba)^{3x} (ba)^{-3} b^3 \\ & = a^3 (ba)^{3x} \cdot e \cdot (ba)^{-3} b^3 \\ & = a^3 (ba)^{3x} \cdot (ba)^{ny} \cdot (ba)^{-3} b^3 \\ & = a^3 (ba)^{3x+ny} ba^{-3} b^3 \\ & = a^3 (ba) (ba)^{-3} b^3 \quad \textcolor{magenta}{\boxed{\text{from \eqref{eq:01Mar2023-1}}}} \\ & = a^3 (ba) \left( a^{-3}b^{-3} \right) b^3 \\ & = a^3 b a^{-2}. \end{align*}

Therefore, we got \begin{align*} & ab = a^3 b a^{-2} = a^3 b^3 b^{-2} a^{-2} \\ \implies & ab = (ab)^3 b^{-2}a^{-2} \\ \implies & (ab)^2 b^{-2} a^{-2} = e\\ \implies & (ab)^2 = a^2 b^2 \\ \implies & ab ab = a^2 b^2 \\ \implies & a^{-1}(ab ab)b^{-1} = a^{-1} \left( a^2 b^2 \right)b^{-1} \\ \implies & ba = ab. \end{align*} Hence, we proved that $G$ is abelian.

Second method (taken from here). Given that for any $a,b\in G$ \begin{align*} (ab)^3 = a^3 b^3 & \implies (ab) (ab) (ab) = a^3 b^3 \\ & \implies a^{-1}ab(ab)ab b^{-1} = a^{-1}a^3b^3 b^{-1} \\ & \implies baba = a^2 b^2 \\ & \implies (ba)^2 = a^2 b^2. \end{align*} Thus, \begin{equation}\label{eq:01Mar2023-2} (ba)^2 = a^2b^2 \qquad\forall~a,b\in G. \end{equation}

Now we consider the commutator of $a,b$ which is $aba^{-1} b^{-1} $ \begin{align*} \left( aba^{-1} b^{-1} \right) ^2 & = \left( a^{-1} b^{-1} \right) ^2 (ab)^2 \quad \textcolor{magenta}{\boxed{\text{from }\eqref{eq:01Mar2023-2}}}\\ & = b^{-2} a^{-2} b^2 a^2 \quad \textcolor{magenta}{\boxed{\text{from }\eqref{eq:01Mar2023-2}}}\\ & = b^{-2} \left( ba^{-1} \right) ^{-2} a^2 \\ & = b^{-2} ba^{-1} ba^{-1} a^2 \\ & = b^{-1} a^{-1} ba. \end{align*} Therefore, \begin{equation}\label{eq:01Mar2023-3} \left( aba^{-1} b^{-1} \right) ^2 = b^{-1} a^{-1} ba \qquad\forall~a,b\in G. \end{equation}

Squaring \eqref{eq:01Mar2023-3}, we get \begin{align*} \left( aba^{-1} b^{-1} \right) ^4 & = \left( b^{-1} a^{-1} ba \right) ^2 \\ & = aba^{-1} b^{-1} \quad \textcolor{magenta}{\boxed{\text{from } \eqref{eq:01Mar2023-3}}} \\ \implies \left( aba^{-1} b^{-1} \right) ^3 & = e. \end{align*} Note that we have given that the order of the group is not divisible by $3$ and hence none of the element order will be $3$. Therefore, \[ aba^{-1} b^{-1} = e \implies ab = ba. \] Hence, $G$ is abelian.