How is the combinatorist's Lagrange Inversion equivalent to the complex analytic expression of the same theorem? I will present three equivalent combinatorical (I assume so, due to the contexts of the papers I found them in) formulations of the Lagrange Inversion Theorem - I have been baffled over the course of many hours as to how they agree with the Wikipedia statement of the theorem! In particular, it has been hinted but never really stated that Wikipedia's theorem is the same as everyone else's - so I don't even know if I'm wasting my time trying to make them match. If anyone knows where the link lies (or whether there is definitely no link) I'd greatly appreciate that - as usual, one has a hard time corroborating Wikipedia's mathematical statements. I did try to parse their "formal residues" explanation, but formal power series are not something I know much about, but their complex-analytic statement is of great interest to me.
In these formulations, $[x^n]\{f(x)\}$ denotes the $n$th coefficient of the curly-braced power series in $x$.

Let $R(t)$ be a power series not involving $x$. Then there is a unique power series $f=f(x)$ such that $f(x)=x\cdot R(f(x))$, and for any Laurent series $\varphi(t)$ not involving $x$, and any integer $n\neq0$, we have: $$\tag{1}[x^n]\{\varphi(f)\}=\frac{1}{n}[t^{n-1}]\{\varphi'(t)\cdot R(t)^n\}$$

From another source:

Suppose $u=u(x)$ is a power series in $x$ satisfying $x=\frac{u}{\varphi(u)}$ where $\varphi(u)$ is a power series in $u$ with a nonzero constant term. Then for any power series $F(u)$ of $u$, and $n(\in\Bbb Z)\neq0$ we have: $$\tag{2}[x^n]\{F(u(x))\}=\frac{1}{n}[u^{n-1}]\{F'(u)\cdot\varphi(u)^n\}$$

I write both presentations in case one is more revealing than the other.
Another strangely put but more promising analytic formulation from the same sources (with proof!):

Let $G(t)=\sum_{n=0}^\infty g_nt^n$, where the $g_i$ are indeterminates. Then there is a unique power series $f$ satisfying: $$f=x+G(f)$$
-Note how this formulation uses an addition, not a multiplication-
Then for any power series $\varphi(t)$ we have: $$\tag{3}\varphi(f)=\varphi(x)+\sum_{m=0}^\infty\frac{1}{m!}\frac{d^{m-1}}{dx^{m-1}}\left(\varphi'(x)\cdot G(x)^m\right)$$

But try as I might, I can think of no sensible choice of $f,G,\varphi,u,R$, whatever, such that this boils down to:

If $f$ is an analytic function in some neighbourhood of $a\in\Bbb C$, such that $f'(a)\ne0$, then (by the inverse function theorem), it has an analytic inverse $g$ in some neighbourhood of $a$, such that if $z=f(w),\,g(z)=w$ for all $w$ in this neighbourhood. The Lagrange Inversion Theorem gives its power series as follows: $$\tag{4}g(z)=a+\sum_{n=0}^\infty\frac{(z-f(a))^n}{n!}\cdot\lim_{w\to a}\frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w-a}{f(w)-f(a)}\right)^n\right]$$

At first, I attempted to show (by uniqueness of power series) that the coefficients in $4$ are in fact the same as $D_z^n g(z)|_{z=f(a)}$, but that went unsuccessfully. I have spent quite some time with $1$ and $2$, attempting to make them play nice (I saw in examples that it was common to let $F$ (as in $2$) equal the identity, so you just get $[x^n]\{u(x)\}=\frac{1}{n}[u^{n-1}]\{\varphi(u)^n\}$), but as the inverse in these combinatorical formulations is weird ($u=x\cdot\varphi(u)$ does not yield the straightforward coefficient in $4$ to my eyes!) I had no luck. I can recognise Wikipedia's coefficient for $g$ as: $$D_z^{n-1}[(g'(z))^n]\big|_{z=f(a)}$$
Which suggests that the $\varphi$ function in $2$ should perhaps be $g'(z)$, but as $\varphi$ has a bizarre relationship I couldn't proceed from here. The only successful observation I have made is that the $\frac{1}{n}$ does indeed make the $n!$ divisors balance - no further progress has been made.
It has been shown to me - here - that $1,2,3$ are equivalent - but how are they equivalent to $4$?
 A: Note that the formula $4$ is not quite equivalent with $1,2,3$ in the sense that the analog of $4$ sometimes called Lagrange Burmann formula in Wikipedia is, as $4$ is more or less just a restatement of $1$ for the case $\phi$ identity, $a=f(a)=0$, while Lagrange Burmann is a restatement of $1$ for general $\phi$
So assuming as above $a=0, f(0)=0, f'(0) \ne 0$ $f(z)=zg(z)$ so $z=f^{-1}(z)g(f^{-1}(z))$ or $h=f^{-1}$ satisfies the first version above with $R(t)=1/g(t)=t/f(t)$
But now it follows by version $1$ applied to $\phi=id$ that the $n$th coefficient of $f^{(-1)}$ is given by $1/$ times the $n-1$th coefficient of $(t/f(t))^n$ which is precisely $\frac{1}{n}\frac{1}{(n-1)!}\lim_{w\to 0}\frac{d^{n-1}}{dw^{n-1}}\left[\left(\frac{w}{f(w)}\right)^n\right]$ so we are done with $1$ (for $\phi$ identity) implies $4$
Conversely we notice that $4$ is essentially $1$ for $\phi$ identity (note that $1$ implies $R$ has non zero constant term as otherwise $f=0$ by induction and the identity is trivial, so with $g(z)=zR(z)$ invertible near zero by the above $f=g^{-1}$ is the given unique power series etc).
