Let $p \in \mathbb N$ be prime, $q \in \mathbb N$ coprime to $p$, and let $F = \mathbb F_p(t)$ the field of rational functions of $t$ with coefficients in $\mathbb F_p$. Consider $$ f(x) = x^{pq} - t. $$ **EDIT** : By Eisenstein's criterion, $x^{pq} - t$ is irreducible over $\mathbb F_p[t]$ (because $t$ is a prime in there). By Gauss' Lemma, it is also irreducible over the field of fractions, which is $\mathbb F_p(t)$. Thanks to Sam L. for this part of my argument.
Since the derivative of $f$ is zero in $\mathbb F_p(t)[x]$, the polynomial is inseparable. But the polynomial $x^q - 1$ is separable in $\mathbb F_p(t)[x]$, because its derivative is $qx^{q-1}$, which has no common roots with $x^q - 1$, so that the roots of $x^q - 1$ are distinct. Now letting $\sqrt[pq]t$ be a root of $x^{pq} - t$ and $w$ a $q^{th}$ root of unity. Then the distinct roots of $f$ are $w^i (\sqrt[pq]t)$, with $i$ ranging from $0$ to $q-1$, each with multiplicity $p$.
Hope that helps,