I was leafing through the "Introduction to Classical Real Analysis" (Stromberg), and while reading the paragraph "Equivalence Relations" in the "Preliminaries" section, I saw:
One checks that any such $R$ [i.e. which is reflexive, symmetric and transitive] satisfies $R^{-1}=R=R \circ R$ and that, conversely, any relation $R$ satisfying these two equalities is an equivalence relation on its domain.
I am not able to prove that $R^{-1}=R=R \circ R$ implies $R$ being reflexive, that is, why if $R^{-1}=R=R \circ R$ then $$\newcommand{\dom}{\operatorname{dom }} x \in \dom R \implies (x,x) \in R \;?$$
Here the domain of the binary relation $R$ over a set $X$ is defined as $$\dom R = \{x \in X : (x,y) \in R \textrm{ for some } y \in X\}.$$
As correctly pointed out by José Carlos Santos, the fact that $R^{-1} = R = R \circ R$ does not imply that $R$ is an equivalence relation over a set $X \neq \emptyset$ in the case of $R = \emptyset$. But it does imply that $R$ is an equivalence relation over the domain of $R$ defined as $\mathrm{dom}(R) = \{x \in X \mid \exists \, y \in X : x \,R\, y\}$ (which is actually what the book you cited claims on p. 3). Let us prove it.
Reflexivity over $\mathrm{dom}(R)$: Let $x \in \mathrm{dom}(R)$. By definition of domain of $R$, there is $y \in X$ such that $x \, R \, y$. Since $R^{-1} = R$, then $y \, R \, x$. Since $R = R \circ R$, from $x \,R\, y$ and $y \,R\, x$ it follows that $x \,R\, x$.
Symmetry: Let $x \,R\, y$. Since $R^{-1} = R$, then $y \,R\, $.
Transitivity: Let $x \,R\, y$ and $y \,R\, z$. Then, $x \,R\, z$ because $R = R \circ R$.
Clearly, if to the hypothesis $R^{-1} = R = R \circ R$ we add the further hypothesis $\mathrm{dom}(R) = X$, then $R$ is an equivalence relation over $X$.
Note that the fact that $R^{-1} = R = R \circ R$ with the further hypothesis $R \neq \emptyset$ does not imply the reflexivity of $R$ over $X$ if $X$ has at least two elements. For instance, take $X = \{0,1\}$ and $R = \{(1,1)\}$.