I don't think I fully understand.
Let's say there is a root $x_0 \in K=\mathbb{F}_p(t)$, where $p$ is a prime number.
Then $x_0 = \frac{P(t)}{Q(t)}$ for some polynomials $P,Q \in \mathbb{F}_p[t]$. We can assume $\gcd(P,Q)=1$
and $x_0^p-t= \frac{(P(t))^p}{(Q(t))^p}-t= \frac{P(t)^p-tQ(t)^p}{Q(t)^p} =0$, so coefficients of $P(t)^p, tQ(t)^p$ must be identical, which contradicts $\gcd(P,Q)=1$,
hence such $x_0$ does not exist and the polynomial has no root in $K$.
am I about right?
anyway, I would appreciate an explanation about $\mathbb{F}_p(t)$, what is $t$? what is the meaning of a variable which does not belong to any specific "world"? I cannot use $t$ as if it was a member of $\mathbb{F}_p$ and hence cannot assume $t^{p-1} = 1 \pmod p$...
The field $K=\mathbb{F}_p(t)$ is the field of fractions of $R=\mathbb{F}_p[t]$, which is a principal ideal domain. Since $x^p-t$ is a polynomial in $R[x]$, Eisenstein's criterion applies and $x^p-t$ is irreducible in $R[x]$, so also in the polynomial ring $K[x]$, by Gauss' lemma.
In particular $x^p-t$ has no roots in $K$.
Can you do it without appealing to Eisenstein's criterion? Yes, of course. A root should be of the form $P(t)/Q(t)$, with $P(t),Q(t)\in R[t]$. Then $$ P(t)^p=tQ(t)^p $$ and so $$ p\deg P(t)=1+p\deg Q(t) $$ which is impossible, because $p$ doesn't divide $1$.
Your text certainly denotes by $t$ an indeterminate, just like $x$, or, equivalently, any element in a field extension of $\mathbb{F}_p$ that is transcendental over $\mathbb{F}_p$.
Indeed, if $t$ is an algebraic element over $\mathbb{F}_p$, the field $\mathbb{F}_p(t)$ is finite, hence perfect, which means that any element has a $p$-th root and the polynomial $x^p-t$ is reducible.