poles of the rational differential form $df$

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I'm working with divisors rational differential forms $\omega\in\Omega^1_{k(C)/k}$ where $C$ is a projective (irreducible) curve over an algebraically closed field $k$. In Silverman, The arithmetic of elliptic curves, Proposition II.4.3(b) is written that if $f\in k(C)$ is regular at $P$ (that is $f\in\mathcal{O}_{C,P}$) then $df/dt$ is also regular at $P$ (where $t$ is an uniformiser, so I guess that $C$ is smooth).

I'm seeking for a proof of that. In Silverman is written to see in Hartshorne the comment following IV.2.1: I don't see the link. There is an other reference that I don't have (Robert, Elliptic curves, II.3.10).

Ideas:

If $f$ is regular then $f=at^n$ with $a$ locally invertible, so $df=nat^{n-1}dt+t^nda$ so $df/dt=nat^{n-1}+t^nda/dt$ so I can suppose that $f\in \mathcal{O}_{C,P}^*$. It doesn't help me.

If $f$ regular then there is an open neighborhood of $U$ of $P$ with $f\in\mathcal{O}_X(U)$ so that $df\in\Omega^1_{X/k}(U)$, so I have to prove that the $\omega\in\Omega^1_{C/k}(U)$ don't have any pole in $U$. That should be evident but how prove it?