= Solution
First let $n\geq0$. The ordinary <Leibniz rule>, applied to a test <function> $h$, gives
$$
\partial^n(fh)=\sum_{r=0}^n\binom nr f^{(r)}h^{(n-r)}.
$$
Equivalently, the composition of <differential operators> is
$$
\boxed{\partial^n m_f=\sum_{r=0}^n\binom nr m_{f^{(r)}}\partial^{n-r}.}
$$
For completeness, this formula follows by <induction>: applying $\partial$ differentiates either the <coefficient> or the remaining <derivative> of $h$, and the two contributions combine by $\binom nr+\binom n{r-1}=\binom{n+1}r$. The initial case is multiplication by $f$.
For all <integers> $n$, define $\binom n0=1$ and $\binom nr=n(n-1)\cdots(n-r+1)/r!$. The <formal pseudodifferential composition rule> is
$$
\boxed{\partial^n m_f=\sum_{r\geq0}\binom nr m_{f^{(r)}}\partial^{n-r},\qquad n\in\mathbb Z.}
$$
For $n=-s<0$, $\binom{-s}r=(-1)^r\binom{s+r-1}r$, so this is usually an infinite series. In particular
$$
\partial^{-1}m_f=m_f\partial^{-1}-m_{f'}\partial^{-2}+m_{f''}\partial^{-3}-\cdots.
$$
Applying $\partial$ on the left gives $m_f$: all later <coefficient> <derivatives> cancel in pairs. This uniquely determines the normal-ordered inverse expansion. Repeating the inverse construction, or using the same binomial recurrence downward in $n$, gives every negative power. More explicitly, applying $\partial$ to the proposed $n$-series yields <coefficient> $\binom nr+\binom n{r-1}=\binom{n+1}r$ at $f^{(r)}\partial^{n+1-r}$, so it is consistent with the already determined $(n+1)$-series. Negative powers here are formal; differentiation on all of $C^\infty(\mathbb R)$ has a nontrivial kernel and has no genuine two-sided inverse there.
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