First of all, $\mathbb{F}_p(t)$, the field of rational functions, has characteristic $p$, since it contains $\mathbb{F}_p$, right?
With that in mind, we can see that $x^p -t$ is not separable over this field, because its formal derivative is $px^{p-1} - 0 = 0$. And it holds for any $p$.
On the other hand, by fixing $p=3$, we can see that the following three values are roots of $x^3 -t$:
$\alpha_0 = t^{1/3}$, because $\alpha_0^3 = t$.
$\alpha_1 = -\sqrt[3]{-1}t^{1/3}$, because $\alpha_1^3 = (-1)^3(\sqrt[3]{-1})^3 (t^{1/3})^3 = -1 \cdot -1 \cdot t = t$.
$\alpha_2 = (-1)^{2/3}t^{1/p}$, because $\alpha_2^3 = ((-1)^{2/3})^3(t^{1/3})^3 = (-1)^2 \cdot t = t$.
Since $x^3 - t$ has three distinct roots, it is separable, isn't? And thus, we have a contradiction.
So, what is the point I'm missing?
By writing "$\sqrt[3]{-1}$" you're assuming that the polynomial $x^3+1$ has a root other than $-1$, but $x^3+1=(x+1)^3$, so the only root is $-1$ and your three roots are all equal.