Solution of functional equation $f(x/f(x)) = 1/f(x)$?

I've been trying to add math rigor to a solution of the functional equation in [1], eq. (22). It is:
$$
f\left(\frac{x}{f(x)}\right) = \frac{1}{f(x)}\,,
$$
where you know that $f(0)=1$ and $f(-x) = f(x)$.

I've been trying to fill in the missing steps. Let's use a substitution $g(x) = \frac{x}{f(x)}$. So we rewrite the functional equation as
$$

x = \frac{\frac{x}{f(x)}}{f\left(\frac{x}{f(x)}\right)}
$$
and get
$$
x = g(g(x))\,.
$$
Then we calculate $g'(0)$ as follows:
$$
g'(0) = \left.\left(\frac{x}{f(x)} \right)'\right|_{x=0}
= \left.\frac{f(x)-xf'(x)}{f^2(x)}\right|_{x=0}

= \frac{f(0)}{f^2(0)}
= 1\,.
$$
Also we can use that $g(x)$ is odd
$$
g(-x) = \frac{-x}{f(-x)} = -\frac{x}{f(x)} = -g(x)\,.
$$
Whenever $g(g(x)) = x$, I have found that this is called involution. That can have many solutions, but using $g(-x)=-g(x)$ and $g'(0)=1$, there should be a way to prove that $g(x) = x$, from which $f(x)=1$ as the only solution.

The article [1] just says, that since $g(x) = g^{-1}(x)$, the graph of $g(x)$ and its inverse must be symmetric along the $y=x$ axis, and since the tangent at $x=0$ is equal to this same line, then $g(x)=x$ must be the only solution. They do mention that this is not completely rigorous proof.

You can assume that $f(x)$ is differentiable. It'd be nice if it can be proven only for continuous functions $f(x)$, but any proof at first would be a good start, even with stronger assumptions.

Update: another idea is to use the fact, that if $a$ and $b$ are two points in the domain of the $g$ function for which $g(a)=g(b)$, then it follows that
$g(g(a)) = g(g(b))$ and $a = b$, which proves that the function is one-to-one.
Given that $g$ is continuous, it means that it is strictly increasing or decreasing [2]. From that, let's say it's increasing, then we can use $y=g(x)$, if $y < x$, then $g(y) < g(x)$ from which $g(g(x)) < y$ and $x < y$ which is a contradiction. Similarly for $y > x$. So we must have $y=x$, from which $g(x)=x$. If $g$ is decreasing, then we set $h(x)=-g(x)$, obtain $h(x)=x$ and $g(x)=-x$. Unless I made a mistake we got $g(x)=\pm x$. And using $g'(0)=1$, we get $g(x)=x$ as the only solution. But I don't feel very good about this proof yet.

[1] Levy-Leblond, J.-M. (1976). One more derivation of the Lorentz transformation. American Journal of Physics, 44(3), 271–277.

[2] A continuous, injective function $f: \mathbb{R} \to \mathbb{R}$ is either strictly increasing or strictly decreasing.