Show that the commutator subgroup $G'$ of a group $G$ is a subgroup of $G^2$ defined by $G^2=\{x^2:x \in G\}$.
I've tried writing the general element of $G'$; $hgh^{-1}g^{-1}$ as the square of some element. Haven't quite found it yet. Any Hints would be appreciated!
The claim is false as written: the subset you have labeled "$G^2$" is not necessarily a subgroup of $G$, so saying that $G'$ is a "subgroup" of something that is not even a group is incorrect. (The smallest examples of groups for which the set of squares is not a subgroup have order $16$, and there are two of them)
But I don't think that's the correct definition of $G^2$ anyway. The usual definition for not-necessarily abelian groups is: for any $n\gt 0$, $$\begin{align*} G^{\{n\}} &= \{g^n\mid g\in G\},\\ G^n &= \langle G^{\{n\}}\rangle = \langle g^n\mid g\in G\rangle. \end{align*}$$
Note that $G^n$ is always normal (in fact, characteristic): because for every $x\in G$ and every element $g^n$ of its generating set, we have $xg^nx^{-1} = (xgx^{-1})^n\in G^{\{n\}}$.
Now consider $G/G^2$ (with the corrected definition). This is a group in which every element satisfies $x^2=e$, and therefore $G/G^2$ is abelian.
But $G'$ has the property that for every normal subgroup $M$ of $G$, if $G/M$ is abelian then $G'\subseteq M$. Therefore, $G'\subseteq G^2$.