Let the formal power series $\phi_1,\dots,\phi_m$ in the variables $x_1,\dots,x_m$ be defined by \begin{equation} \phi_i(x_1,\dots,x_m)=x_i\rho_i(\phi_1,\dots,\phi_m),\qquad i = 1,\dots,m \end{equation} for some formal series $\rho_i(x_1,\dots,x_m)$. Then, for any formal power series $f(x_1,\dots,x_m)$ we have \begin{equation} [{\vec x}^{\,\vec n}]f(\vec \phi(\vec x)) = [{\vec t}^{\,\vec n}]f(\vec t) \text{det}[K(\vec t)] {\vec \rho}^{\,\vec n}(\vec t) \end{equation} where $K(\vec t)$ is a matrix from $\mathbb R^{m\times m}$, \begin{equation} K(\vec t)_{i,j}=\delta_{i,j}-\frac{t_i}{\vec \rho_i(\vec t)}\frac{\partial \vec \rho_i}{\partial t_j}(\vec t), \qquad i,j\in 1,\dots,m \end{equation} where $\vec t=(t_1,\dots,t_m)$, $\vec n =(n_1,\dots,n_m)$, ${\vec x}^{\,\vec n}=x_1^{n_1}x_2^{n_2}\cdots x_m^{n_m}$ and $\vec x(\vec y)=[x_1(\vec y),\dots,x_m(\vec y)]$.
In the case that $m=1$, we recover the standard Lagrange-inversion formula.
Consider the system \begin{equation} \phi_i(x_1,\dots,x_m)=x_i\rho_i(\phi_1^{p_1},\dots,\phi_m^{p_m}),\qquad i = 1,\dots,m \end{equation} where $p_i$ are integer powers. How can I apply the Lagrange inversion theorem in this case?
For a 1-dimensional scenario, my attempt is as follows. The coefficient of $x^n$ in $f(\phi^m(x))$ with $\phi(x) =x\rho(\phi^m(x))$ for $m=1,2,\dots$. I find \begin{align} [x^n]f(\phi^m(x))=&\frac{n!}{2\pi i}\oint \frac{f(\phi^m(x))}{x^{n+1}}dx\\ =& \frac{(n-1)!}{2\pi i}\oint \frac{1}{x^n}\left(\frac{d}{dz}f(\phi^m(x))\right)dz\\ =& \frac{(n-1)!}{2\pi i}\oint \frac{1}{x^n}\left(f'(\phi^m(x))m\phi^{m-1}(x)\phi'(x)\right)dx\\ =&\frac{m(n-1)!}{2\pi i}\oint\frac{ \rho(\phi^m(x))^n}{\phi^n(x)}\left(f'(\phi^m(x))\phi^{m-1}(x)\phi'(x)\right)dx \end{align} In the last step we considered $x$ as a function of $\phi$ and $\rho$ from above. We now change the volume element and move the $\phi^{m-1}$ term to the denominator to obtain \begin{equation} [x^n]f(\phi^m(x))=\frac{m(n-1)!}{2\pi i}\oint\frac{ 1}{\phi(x)^{n-m+1}}\left[\rho(\phi^m(x))^nf'(\phi^m(x))\right]d\phi \end{equation} We must then undo the Cauchy formula; however, this is where I am a bit stuck due to the denominator in $\phi(x)$. My attempt is as follows $$ [x^n]f(\phi^m(x))= \frac{m}{n}[t^{n-m}]\rho(t^m)^nf'(t^m) $$ where we have again simplified $1/n=(n-1)!/n!$. I am not certain this is correct; however.
