Here's what I've got, but I'm not confident it's airtight...
Let $K=\mathbb F_p(X^6)$ then $\mathbb F_p(X)=K(X)$ and the polynomial $F(T) = T^6-X^6\in K[T]$ has $X$ as a root. It's irreducible by Eisentstein's crieterion taking the irreducible element $P=X^6\in\mathbb F_p[X^6]$, since $P\mid X^6$ and $P^2\nmid X^6$, so we know $[K(X):K]=\deg F=6$.
There's a sufficient criterion of separability that if $\operatorname{char}K=p$ and $p\nmid [L:K]$ then $L/K$ is separable. Since $\mathbb F_p\subset K$, $\operatorname{char}K=p$ and so $K(X)/K$ is separable for all $p\nmid 6$, i.e: $p\ne 2,3$.
Since this criterion is only sufficient, we need to verify that $K(X)/K$ is inseparable for $p=2,3$. In either case, $F$ contains mulitple roots, and so the extension cannot be separable:
- In $\mathbb F_2(X)[T]$, $T^6-X^6=(T^3-X^3)^2$, so each root has multiplicity $2$
- In $\mathbb F_3(X)[T]$, $T^6-X^6=(T^2-X^2)^3$, so each root has multiplicity $3$
Thus: $\mathbb F_p(X)/\mathbb F_p(X^6)$ is separable for all primes $p \geq5$.
This solution seemed ... so simple that I feel like I've missed something. Is it OK?