Let $K$ be any field and $L=K((1/t))$ be the field of Laurent series over $K$.
My question is: given a rational function $D\in K(t)$ and a natural number $v$, when is $D^{1/v}$ an element of $L$?
In particular, is it true that for $K=\mathbf{F}_q$, a finite field, and $v\mid (q-1)$, we can always find a Laurent series expansion for $D^{1/v}$, $D$ a monic polynomial?
Let $\mathrm{ord}$ be the discrete valuation on $L$ (written additively) such that $\mathrm{ord}\,t =1$, etc. Let $f$ be a nonzero element of $L$. If $f^{1/v}$ makes sense, then $v\,\mathrm{ord}\,f^{1/v}=\,\mathrm{ord}\,f$, so a necessary condition for a $v$th root of some nonzero $f:L$ to exist is that $v$ divides $\mathrm{ord}\,f$. Pulling out a factor of $t^{\mathrm{ord}\,f}$ we may assume without loss of generality that $\mathrm{ord}\, f=0$.
If $\mathrm{ord}\,f=0$, then $f^{1/v}$, if it makes sense, must satisfy $f^{1/v}(0)^v=f(0)$, so $f(0)$ must be a $v$th power in $K$.
Conversely, if $v$ divides $\mathrm{ord}\,f$, pull out a factor of $t^{\mathrm{ord}\,f}$ so that we may assume that $\mathrm{ord}\,f=0$. If the polynomial $x^v-f(0)$ admits a simple root in $K$ (e.g., if $v$ and the characteristic of $K$ are coprime and $f(0)$ is a $v$th order residue), then Hensel's lemma guarantees for each such root a corresponding root of $x^v-f$ in $L$.
Example: let $f$ be a polynomial with coefficients in $\mathbf{F}_p$ ($p\neq 2$) such that $f(0)\neq 0$. Then $f$ admits a square root in $L$ if and only if $f(0)$ is a quadratic residue modulo $p$.