How does this order-by-order inversion of series done? In the paper: DOI: 10.1103/PhysRevE.99.022411, on page 3, it is given that

$$
\psi^{\prime}(s)=\sum_{m=0}^{\infty} \frac{2 \mathcal{B}_{2 m}}{(2 m) !} \frac{\phi^{(2 m)}(s)}{\ell_{0}^{2 m-1}}
$$
wherein $\mathcal{B}_{0}=1, \mathcal{B}_{1}=-\frac{1}{2}, \ldots$ are the Bernoulli numbers (of the first kind).... While we are not aware of any explicit expression for the coefficients of the inverted series, it is straightforward to invert the series order-by-order by substituting back and forth, and thus obtain
$$\phi(s)=\frac{\psi^{\prime}(s)}{2 \ell_{0}}-\frac{\psi^{\prime \prime \prime}(s)}{24 \ell_{0}^{3}}+\frac{\psi^{(\mathrm{v})}(s)}{240 \ell_{0}^{5}}+\cdots$$

How is this order-by-order inversion of series done?
 A: Suppose it is given that
$$ \psi^{\prime}(s)=\sum_{m=0}^{\infty} \frac{2 B_{2 m}}{(2 m) !} \frac{\phi^{(2 m)}(s)}{\ell_{0}^{2 m-1}} = 2\ell_0 \phi(s) + \frac{\phi''(s)}{6\ell_0} -\frac{\phi^{(4)}(x)}{360\ell_0^3} + \dots $$
where $\,B_{2m}\,$ the the Bernoulli numbers and $\,\ell_0\,$
does not depend on $\,s.\,$ To find $\,\phi(s)\,$ in
terms of $\,\psi\,$ and its derivatives a natural Ansatz is
$$ \phi(s) = a_1 \frac{\psi'(s)}{1!\,\ell_0} +
 a_3 \frac{\psi^{(3)}(s)}{3!\,\ell_0^3} + 
a_5 \frac{\psi^{(5)}(s)}{5!\,\ell_0^5} + \dots $$
for some rational coefficients $\,a_n.\,$
Use the first equation to find the odd derivatives
of $\,\psi(s)\,$ and substitute them in the Ansatz
and compare with $\,\phi(s).\,$ This gives the
linear equations
$$ 0 \!=\! -\!1\!+\!2a_1 \!=\! a_1\!+\!2a_3 \!=\! 
a_1\!-\!10a_3\!-\!6a_5 \!=\! a_1\!-\!7a_3\!+\!21a_5\!+\!6a_7
 =\dots.$$
Solve the linear equations one by one to get
$$ a_1 = \frac12,\,a_3 = -\frac14,\,a_5 = \frac36,\,
a_7 = -\frac{17}8,\, a_9 = \frac{155}{10},\,\dots. $$
The coefficients are explicitly given by
$$ a_n = \frac{B_{n+1}}{n+1}(2^{n+2}-2).$$
Note that $\,-(n+1)a_n\,$ is OEIS sequence A001469 "Genocchi numbers (of the first kind)".
