Greeting, i'm trying to solve the following exercise of Q.Liu algebraic geometry and arithmetic curves
My attempt: Since every rational point is contained in some affine scheme, we immediately reduce to the case where $X=\mathrm{spec} A$ is affine, therefore we might suppose as well that $O_X(X)=A$ and that $f=\varphi^{-1}$, let $p\in \mathrm{spec} A$ a $k$-rational point. Fix an isomorphism $k(x)\simeq k$. It's not hard to see that
$P(X_1,\dots,X_n)\in \varphi^{-1}(p) \Leftrightarrow P(f_1(x),\dots,f_n(x))=0 \Leftrightarrow P\in (X_1-f_1(x),\dots,X_n-f_n(x))$
However i have my doubts about this solution because $f_1(x),\dots,f_n(x)$ depend on the choice of the isomorphism $k(x)\simeq k$.
I'll appreciate any help, thanks in advance.
This sort of confusion around "an isomorphism $k(x)\cong k$" for $x$ a $k$-rational point on a $k$-scheme is somewhat common and has appeared on this website before: here, here, and here for example. The key point is that when someone says "let $x$ be a $k$-rational point inside a $k$-scheme $X$", they mean that the induced map on residue fields $k\to k(x)$ coming from the map $\mathcal{O}_{\operatorname{Spec} k,(0)}\to\mathcal{O}_{X,x}$ is an isomorphism, or that $k\cong k(x)$ as $k$-algebras, which is stronger than just as fields.
This resolves your issue because it removes the choice you're worried about with the isomorphism $k(x)\cong k$. The solution you've come up with is the right idea (though I confess I'm a little unsure about what you've done with letting $f=\varphi^{-1}$ - is this supposed to be something about the map on ideals, or maybe you're after the induced map on Spec?).
Alternatively, one can consider the map $\operatorname{Spec} k\to X$ picking out your $k$-rational point $x$ and the composition with $X\to \Bbb A^n_k$: after taking global sections, we get a map $k[T_1,\cdots,T_N]\to \mathcal{O}_X(X) \to k$, and the composition $k[T_1,\cdots,T_n]\to k$ exactly tells you what the maximal ideal of $k[T_1,\cdots,T_n]$ cutting out the image of our $k$-rational point is: it's $(T_1-f_1(x),\cdots,T_n-f_n(x))$, so $(0)\in\operatorname{Spec} k$ maps to $(T_1-f_1(x),\cdots,T_n-f_n(x))\in\operatorname{Spec} k[T_1,\cdots,T_n]$, which is identified with $(f_1(x),\cdots,f_n(x))$ under the identification $\Bbb A^n_k(k)=k^n$.