If $F/K$ is a field extension and $f$ separable over $F $ then $ f$ is separable over $K$ The following is a test question which I found to be suspiciously obvious to be 
asked in an exam: 

Let $F/K$ be a field extension (no hypotheses over it) and $f \in
 K[X]$ then:
$f$  separable over $F \implies f$ is separable over $K$.

I would reason that since the irreducible factors over $F$ should have all its roots simple, then the same must happen for it over $K$. But the fact that there is no hypotheses over the extension confuses me.
Is my approach correct?
 A: $\newcommand{\FF}{\mathbb{F}}\newcommand{\tens}{\otimes}\newcommand{\mm}{\mathfrak{m}}\newcommand{\into}{\hookrightarrow}$
It is not necessarily obvious, in fact the truth of the statement depends on the definition of separability used. The ease of proof of the statement when true then depends on the characterizations of separability that you are familiar with.
The modern definition of separability is that $f$ is separable over $k$ if it has no repeated roots over any extension field of $k$. Then if $f$ is separable over $F$, but has repeated roots in a field extension $L$ of $K$, then taking the compositum of $L$ and $F$ over $K$, we get a field extension of $F$ containing $L$, hence in which $f$ has repeated roots, contradicting separability of $f$ over $F$. 
Alternatively, $f$ is separable over any field if and only if the resultant of $f$ and $f'$ is nonzero (i.e., if $f$ and $f'$ are relatively prime), which is a polynomial condition on the coefficients, and hence independent of the base field.
However, there is an older definition of separability, that you appear to be referencing in your question: $f$ is separable if all of its irreducible factors are separable in the sense above. With that definition, the statement is false.
Let $K=\FF_p(T^p)$, $F=\FF_p(T)$. Then let $f=X^p-T^p$. $f$ factors in $F[X]$ as $(X-T)^p$, which is separable by this definition, since its irreducible factors, $X-T$, are all separable, but $f$ is not separable over $K$, since $f$ is irreducible over $K$, but its roots are not distinct in $F$. 
Notes:
For both definitions and more information, see wiki.
Also to produce the compositum in general, a priori we may not have $L$ and $F$ all contained in some common field $\Omega$, so we first need to find some such field in order to show that the compositum over $K$ exists. To this end, take $\Omega=(L\tens_K F)/\mm$ where $\mm$ is any maximal ideal of $L\tens_K F$. $\Omega$ is a field by definition, and it has natural maps $L\into \Omega$ and $F\into \Omega$ (since ring homomorphisms of fields are always injective) such that $K\into L \into\Omega = K\into F\into \Omega$. 
