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$1)$ $F=\Bbb F_2(x)$ and consider $K=F(x^{1/6})$. Now $K/F$ is neither normal nor separable.
$2)$ Let $k$ be field of characteristic $2$, let $F=k(x,y)$, let $S=F(u)$, where $u$ is a root of $t^2+t+x$, and let $K=S(\sqrt{uy})$ then $K/F$ is not separable I have proved. How to prove $K/F$ is not normal? I think considering $min(\sqrt{uy},F)$ I have to show this does not split over $K$. But how to prove?
$3$ What is the $3$rd example?
We can construct an example like the one you have given: $${F}_{3}(\sqrt[15]{\mathrm t})/{{F}_{3}(\mathrm t)}\qquad \text { is neither normal nor separable.}$$ In the same manner,actually we can construct as many examples as you want,namely: $${F}_{p}(\sqrt[pq]{\mathrm t})/{{F}_{p}(\mathrm t)}\qquad \text { is neither normal nor separable.}$$ where $p$, and $q$ are two different primes.I will proof the general case in two steps.First I will proof the minimal polynomial of $\sqrt[pq]{\mathrm t}$ over ${ {F}_{p}(\mathrm t)}$ is $ f(x)=x^{pq}-t $;then I will show that $ f(x)=x^{pq}-t $ doesn't split over ${F}_{p}(\sqrt[pq]{\mathrm t}).$
(1)Note that ${F}_{p}(\mathrm t)$ is the fractional field of $ {F}_{p}[\mathrm t]$,which is a $UFD$. $t$ is a prime element on $ {F}_{p}[\mathrm t]$,so $ f(x)=x^{pq}-t $ is irreducible by Eisenstein Criterion .
(2) If we suppose $ f(x)=x^{pq}-t $ does split over $ {F}_{p}(\sqrt[pq]{\mathrm t})$,it is easy to see that this will imply that $ {F}_{p}(\sqrt[pq]{\mathrm t})$ contains all the roots of $ g(x)=x^{pq}-1.$ ( note that all the other roots of $f(x)$ has the form $\zeta^{i}\sqrt[pq]{\mathrm t}$ ,where $ \zeta $ is the $pq \,\text{th}$ primitive root. So $ \zeta $ will lie in $ {F}_{p}(\sqrt[pq]{\mathrm t})$.Suppose $$\zeta =\frac{u(\sqrt[pq]{\mathrm t})}{v(\sqrt[pq]{\mathrm t})},\qquad u,v\in {F}_{p} [x]$$ So we have $$ {\left(\frac{u(\sqrt[pq]{\mathrm t})}{v(\sqrt[pq]{\mathrm t})}\right)}^{pq}-1=0$$ This forces $$(u(\sqrt[pq]{\mathrm t}))^{pq}-(v(\sqrt[pq]{\mathrm t}))^{pq}=0.$$ contradicts to the fact that $\sqrt[pq]{\mathrm t}$ is transcendent over $ {F}_{p}$. Any question or mistake about my proof ,please add your comment.
Additional remarks
I forgot to proof the extension is not separable,which is not to hard: Since $\sqrt[p]{t}\in {F}_{p}(\sqrt[pq]{\mathrm t}) $,$\sqrt[p]{t}$ is a root of $x^{p}-t=0$,but $x^{p}-t=[x-(\sqrt[p]{\mathrm t})]^{p}$ over an extension of $ {F}_{p}(t)$. So the minimal polynomial of $\sqrt[p]{t} $ has only one root,and $\sqrt[p]{t} $ is not separable over $ {F}_{p}(t)$.Our conclusion follows.