Inverse of a composite function

A standard property of composite functions is that: $$(f\circ g)^{-1}(z)=(g^{-1}\circ f^{-1})(z)$$

Is it acceptable to prove this property at the level of a high school proof (not college level), by showing that the LHS of the above is equal to the RHS. For example, here is an attempt below: Define: $$x=g^{-1}(y), y=f^{-1}(z) $$

From the above two we have: $$(f\circ g)(x)=f(y)=z$$ And we can invert the above to get: $$LHS=(f\circ g)^{-1}(z)=x$$ And we can also simplify the right hand side: $$RHS=(g^{-1}\circ f^{-1})(z)=g^{-1}(y)=x$$ Hence, $$LHS=RHS\implies Proved$$

I know the standard proof where I show that: $$(f\circ g) \circ(g^{-1}\circ f^{-1})=\text{Identity Function}$$ $$(g^{-1}\circ f^{-1})\circ(f\circ g) =\text{Identity Function}$$

I am NOT looking for the standard proof.

Maybe I shall just write out a clean proof by elements instead of writing a paramount number of comments.

Lemma. Let $f:X \to Y$ and $g:Y \to X$ be invertible functions between sets. Then, $(f \circ g)^{-1} = g^{-1} \circ f^{-1}$.

Proof. Let $y \in Y$. Then, we write $x = f^{-1}(y)$ and $w = g^{-1}(x)$. In particular, we have $(f \circ g)(w) = f(g(w)) = f(x) = y$ which implies $w = (f \circ g)^{-1}(y)$. Thus, $$ (f \circ g)^{-1}(y) = w = g^{-1}(x) = g^{-1}(f^{-1}(y)) = (g^{-1} \circ f^{-1})(y).$$ Since $y$ was chosen arbitrary, this implies the desired. $\square$

Let me follow with some remarks but these will probably still be a bit too much for you. A good time to come back to this could be when you enter graduate school (or perhaps even earlier). But maybe other readers will appreciate it, so I will leave it here.

But if you scrutinize the proof long enough, this is no more than what you call the "standard proof", only plastering the steps with a lot of element chasing. And as remarked in the comments, your "standard proof" is preferred since it generalizes far more than this proof, namely to all categories. As an example $(gh)^{-1} = h^{-1}g^{-1}$ in groups follows readily from this.

And for the even greater experts, the proof by elements actually also generalizes to the categorical setting by introducing generalized elements. Then, one lands in the viewpoint of the Yoneda Lemma where elements are no more than morphisms.