Let $k^p = k$, $K = k(x)$ and $F = k(u)$ for some $u \in K$. Show that $K/F$ is separable $\iff u \not\in K^p$. 
Let $K$ be the rational function field $k(x)$ over a perfect field $k$ of
  characteristic $p > 0$. Let $F = k(u)$ for some $u\in K$, and write
  $u = f(x)/g(x)$ with $f$ and $g$ relatively prime. Show that $K/F$ is
  a separable extension if and only if $u\notin K^p$.

There is already a post on this but it has not yet been answered : Patrick Morandi- Field and Galois Theory- Section 4- Exercise 11
The truth is that I am very confused with this problem, I would like someone please help me to understand what happens with this exercise and explain to me how it could be done by giving me a help or something. I already have an approach for the left-to-right implication, but I do not know very well how good it is. Could someone help me with the other implication please? Thank you very much.
 A: Try this approach:
Consider the minimal $F$-polynomial for $x$, call it $H(T)$. It’s $F$-irreducible, and so of form $h(T^q)$ with $q=p^s$ for an $s\ge0$ and $h(T)$ an $F$-irreducible polynomial that’s separable.
If $s>0$, then $x$ is not separable over $F$ (and so $K$ is not separable over $F$), but $x^q$ is separable over $F$, as a root of the separable polynomial $h(T)$. We have $F\subset F(x^q)=K^q\subset F(x^p)=K^p\subset F(x)=K$. The last (rightmost) inclusion is strict, since $s>0$, and so $u\in K^p$.
On the other hand, if $s=0$, then $x$ is separable over $F$, the extension $K\supset F$ is separable, and $F\not\subset K^p=k(x^p)$: if we had had $F\subset k(x^p)$, then the minimal $F$-polynomial for $x$ as above would have $s>0$. Since $F\not\subset K^p$, it follows that $u\notin K^p$, since $u\in K^p\Rightarrow k(u)=F\subset K^p$.
So there are two mutually exclusive cases: $s>0$ with $K$ not separable over $F$ and $u\in K^p$; and $s=0$ with $K$ separable over $F$ and $u\notin K^p$.
This argument is not as neat and clear as I’d have liked. Maybe someone can do better.
