This was originally posted on MSE, but after a fair amount of time and a bounty it got no response. Unfortunately I have not yet resolved my doubts.
There's a fair bit of setup here, but you don't necessarily need to read all of it to understand the problem. All notation is as in Hartshorne chapter V section 2 on ruled surfaces.
Setup. Let $k$ be an algebraically closed field of characteristic 3. Let $C=V(x^3y+y^3z+z^3x)\subset\Bbb P^2_k$. Let $X\to C$ be a ruled surface defined as $\Bbb P(\mathscr{E})$ where $\mathscr{E}$ is an extension $$0\to\mathcal{O}_C\to\mathscr{E}\to\mathscr{L}_C(P)\to 0$$ defined by an element $\xi\in H^1(\mathscr{L}_C(-P))$ which pulls back to $0\in H^1(\mathscr{L}(-3P))$ under the $k$-linear Frobenius on $C$ (previous parts of the question are about showing such a thing exists, and that's not an issue for me).
Exercise. Show that $X$ contains a nonsingular curve $Y\equiv 3C_0-3f$, such that $Y\to C$ is purely inseparable.
Question. I'm running in to some issues trying to solve this - I don't think there's a curve that does exactly this. Here's why:
First, by proposition IV.2.5, if we have a finite morphism of smooth projective curves which induces a purely inseparable field extension, then the genera of the two curves are equal. (Stacks 0CD0 confirms this.) So $g(Y)$ and $g(C)$ should be equal here.
Now let's calculate $g(Y)$ by adjunction. Adjunction says that $2g(Y)-2=Y.(Y+K_X)$ (Hartshorne proposition V.1.5). By linearity of the intersection pairing, $2g(Y)-2$ should be divisible by 3, since $Y\equiv 3(C_0-f)$. If $g(Y)=g(C)$ by the above, we know $g(C)=3$ by the degree-genus formula for plane curves, so we should have $2g(Y)-2=4$, which is not divisible by 3.
(In fact, using corollary V.2.11 which says $K_X\equiv -2C_0+(2g(C)-2-e)f$, I get that no curve $Y\equiv 3C_0+bf$ can have genus $3$ as $b$ varies.)
Did I make a mistake here? If so, where? Or does this exercise need some adjustment?
