Do we have $[E : K]_i = [E_r : K]$ where $E_r$ is purely inseparable closure?
Solution 1:
$k=\Bbb{F}_3,K=k(x^6,y^6),E=K(x+y^2),L=K(x^3)=K(x^3+y^6)$.
$L/K$ is a separable quadratic extension, $E/L$ is purely inseparable of degree $3$ and $[E:K]=6$.
If there is $E/F/K$ with $F/K$ purely inseparable then $[F:K]=3,[E:F]=2$.
Let $f\in F[T]$ be the degree 2 minimal polynomial of $x+y^2$.
$f$ stays irreducible over $k(x^2,y^2)\supset F$ so in fact $f=(T-x-y^2)(T+x-y^2)= T^2-2y^2 T-x^2+y^4$ whence $y^2,x^2\in F$ which is a contradiction as $k(x^2,y^2)$ has degree $3^2$ over $K$.