From F. W. Lawvere and S. H. Schanuel's Conceptual Mathematics, 2nd Ed. "Article II: Exercise 10" for self-study.
If $f: A \rightarrow B $ and $g: B \rightarrow C$ are isomorphisms, then $g \circ f$ is an isomorphism too and $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
My attempt is below. It seems too verbose, but I believe I demonstrated the equivalence given the preconditions. Is there a clearer or more insightful way to demonstrate this equivalence?
Proof:
Let an isomorphism be defined for a map $f: A \rightarrow B$ and its inverse $f^{-1} : B \rightarrow A$ where $f \circ f^{-1} = 1_B$ and $f^{-1} \circ f = 1_A$.
Now, allow the same for $g: B \rightarrow C$ and its inverse $g^{-1} : C \rightarrow B$ where $g \circ g^{-1} = 1_C$ and $g^{-1} \circ g = 1_B$.
Thus, $g \circ f : A \rightarrow B \rightarrow C$ and $g^{-1} \circ (g \circ f) : A \rightarrow B \rightarrow C \rightarrow B$ or equivalently $g^{-1} \circ (g \circ f) : A \rightarrow B$ because $1_B \circ f = f$
Further, $f^{-1} \circ g^{-1} \circ (g \circ f) : A \rightarrow B \rightarrow C \rightarrow B \rightarrow A$ and so $f^{-1} \circ 1_B \circ f = f^{-1} \circ f = 1_A$ or equivalently, $f^{-1} \circ 1_B \circ f = f^{-1} \circ f : A \rightarrow A = 1_A$
Now, since we have shown $g \circ f : A \rightarrow C$, by definition of an inverse, we know $(g \circ f)^{-1} : C \rightarrow A$ and further $(g \circ f)^{-1} \circ (g \circ f) : A \rightarrow C \rightarrow A = 1_A$
Finally, since $f^{-1} \circ 1_B \circ f = f^{-1} \circ f = 1_A$ and $(g \circ f)^{-1} \circ (g \circ f) = 1_A$, by transitivity $(g \circ f)^{-1}=f^{-1} \circ g^{-1}$.
A more conceptual way to understand the equivalence is to combine the proof of Proposition 3."If $f\colon A\to B$ has a retraction and if $g\colon B\to C$ has a retraction, then $g\circ f\colon A\to C$ has a retraction", with Theorem (uniqueness of inverses): "If $f$ has both a retraction $r$ and a section $s$, then $r=s$".
A concise proof of the proposition is the following. Suppose we are given $r_1\colon A_0\leftrightarrows A_1\colon s_1$ and $r_2\colon A_1\leftrightarrows A_2\colon s_2$ with $r_i\circ s_i=1_{A_i}$, i.e. with $s_i$ sections of $r_i$, or equivalently $r_i$ retractions of $s_i$.
Then associativity of composition and the identity property of $1_{A_1}$ implies $(r_2\circ r_1)\circ(s_1\circ s_2)=r_2\circ(r_1\circ s_1)\circ s_2=r_2\circ 1_{A_1}\circ s_2=r_2\circ s_2=1_{A_2}$. In other words, $s_1\circ s_2$ is a section of $r_2\circ r_1$, or equivalently, $r_2\circ r_1$ is a retraction of $s_1\circ s_2$.
(Note the above proof of Proposition 3 also solves Exercise 8:"Prove that the composite of two maps, each having sections, has itself a section.")
It now follows that if $f\colon A\to B$ and $g\colon B\to A$ are isomorphisms, i.e. each has both a section and a retraction, then their composite also has both a section and retraction, i.e. is an isomorphism. Moreover, if the (necessarily same by the Theorem) section and retractions of $f$ and $g$ are respectively $f^{-1}$ and $g^{-1}$, then $f^{-1}\circ g^{-1}$ is the corresponding section and retraction of $g\circ f$.