I can recall that an extension $E$ over $F$ is separable if the minimal polynomial of every element of $E$ over $F$ is a separable polynomial.
Further $$[E:F]_s=|\{f:E \to F^a: f \text{ is an } F \text{ embedding}\}$$
If $E$ over $F$ is separable then $[E:F]_s=[E:F]$.
On the other hand, $[E:F]_s=1$ then $E$ over $F$ is purely inseparable.
Now I am searching for examples of $E$ over $F$ such that $E$ is neither separable nor purely inseparable.
I thought of taking $K$ to be a field of $\operatorname{char}(K)=p$ and then consider $K(x^p,y^p) \subset K(x,y)$ and this is not separable for sure but its purely inseparable too. I am not sure but it seems every extension is either separable or purely inseparable.
Best Answer
$\newcommand{\Fp}{\mathbb{F}_p}$You could just do this in two steps:
$$\Fp(x^p) \subset \Fp(x) \subset \Fp(x^{1/2}),$$
where $p$ is any prime $\neq 2$.
Note that separable and purely inseparable extensions both share the "transitive" property, i.e., if you have extensions $$F \subset L \subset K,$$ then $F \subset K$ is (purely in)separable iff both $F \subset L$ and $L \subset K$ are so.
In the example above, the first extension is not separable, whereas the latter is not purely inseparable.