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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?

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    Yes $\mathbb{F}_p$ is the subfield of $K= \mathbb{F}_p(t)$ generated by $1$, so that $\underbrace{1+\ldots+1}_p = p = 0$, thus $\forall a \in K, \underbrace{a+\ldots+a}_p = p a = 0$ i.e. $K$ has characteristic $p$ – 2017-02-12

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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.

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    So, should I be able to write $x^3 - t$ as $(x - \alpha)^3$ or $(x - \alpha)(x - \beta)^2$? – 2017-02-12
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    Yes, it does hold that $x^3-t=(x-\alpha)^3$ in $\mathbb{F}_p(t)(\alpha)$. This follows from the fact that taking the third power is a homomorphism in characteristic 3. – 2017-02-12
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    Yes, sure, I just have to take $\alpha = \sqrt[3]{t}$. Ok. I understand now. I can't just take new elements (like $\sqrt[3]{-1}$) and put them into the field to construct a extension without checking that those new elements are really new (in this case, it is like $\sqrt[3]{-1}$ was just $1$). – 2017-02-12