Is there an alternative method to prove
$(ab)^{2}=a^{2}b^{2}$ for all elements $a,b \in G \implies$ $(ab)^{-1}=a^{-1}b^{-1}$ for all elements $a,b \in G$.
then the one I give below?
Let $G$ be a group and suppose $(ab)^{2}=a^{2}b^{2}$ for all elements $a,b \in G$.
Since $G$ is a group, both $a$ and $b$ have inverses, denoted by $a^{-1}$ and $b^{-1}$, respectively. Multiplication in $G$ is well defined, which mean we can multiply both sides of the equation \begin{align*} a^{-1}((ab)(ab))&=a^{-1}(a^{2}b^{2}) \\ (a^{-1}(ab))(ab)&=(a^{-1}a^{2})b^{2} \\ ((a^{-1}a)b)(ab) &=((a^{-1}a)a)b^{2} \\ (eb)(ab)&=(ea)b^{2} \\ b(ab)&=ab^{2} \end{align*} Furthermore, we may multiply on the right: \begin{align*} (b(ab))b^{-1}&=(ab^{2})b^{-1} \\ ba(bb^{-1})&=ab(bb^{-1}) \\ bae &=abe \\ ba&=ab \end{align*} As $G$ is a group, we have, by proposition $3.1.3$ (*Abstract Algebra * by Beachy & Blair, 91), $(ab)^{-1}=b^{-1}a^{-1}$. So, taking the inverse on both sides we arrive have \begin{equation} a^{-1}b^{-1}=b^{-1}a^{-1}, \quad \forall a,b \in G \end{equation} which proves $(ab)^{-1}=a^{-1}b^{-1}$.