In Hartshorne book Proposition (I.3.3) is that
Proposition : Let $U_i\subset \mathbb{P}^n$ be the open set defined by the equation $x_i\neq 0.$ Then the mapping $\varphi_i : U_i\longrightarrow \mathbb A^n$ is an isomorphism of varieties.
Let $$\varphi_i : U_i\longrightarrow \mathbb A^n\\(a_0 : \cdots : a_n)\longrightarrow (\frac{a_0}{a_i} , \dots ,\frac{a_{i-1}}{a_i},\frac{a_{i+1}}{ a_i},\dots,\frac{a_n}{a_i})$$
$$\varphi_i^{-1} :\mathbb A^n \longrightarrow U_i\\(a_1\dots,a_n)\longmapsto (a_1:\cdots:a_{i-1}:1:a_{i+1}:\cdots:a_n)$$
and
$$\alpha_i : A^h \longrightarrow R \hspace{3.2cm} f \longmapsto f(y_1,\dots,y_{i-1},1,y_i,\dots,y_n) $$
$$\beta_i : R \longrightarrow A^h \hspace{5mm} g \longmapsto x_{i}^{d} g(x_0/x_i,\dots,x_{i-1}/x_{i},x_{i+1}/x_i,\dots,x_{n}/x_{i}).$$
where $R=k[y_1,\dots,y_n],$ $A=k|x_0,\dots,x_n]$ and $A^h$ is the set of homogeneous elements of $A.$
My question is : How to prove that for every open sets $V\subseteq U_i,$ $U\subseteq \mathbb{A}^n$ and for every regulars functions $f: V\longrightarrow k,$ $g: U\longrightarrow k$ the functions $fo\varphi_i:\varphi_i^{-1}(V)\longrightarrow k$ and $go\varphi_i^{-1}:\varphi_i(U)\longrightarrow k$ are regulars using the maps $\alpha_i$ and $\beta_i.$
Any help would be appreciated.
Let me just drop the indices and assume that $i = 0$. I'll just try to get you started.
You give me a regular function $f$ on $V \subseteq \mathbb{A}^n$. I get a function $f \circ \varphi$ on the open set (I think he verifies the continuity in I.2) $\varphi^{-1}(V)$. Pick a point $P$ in that open set. By the regularity of $f$ there is some neighborhood $W$ of $\varphi(P)$ contained in $V$ and $g, h \in R$ such that $f = g/h$ on $W$. You'd like to use these to describe $f \circ \varphi$ on $\varphi^{-1}(W)$.
Well, what else can you do? Shove them into $\beta$. The resulting homogeneous polynomials may not have the same degree, so multiply one of them by a power of $x_0$ if you must (this part might seem a little unnatural; it doesn't come up in the other direction). Call the results $g', h'$. If I haven't fouled this up, you should be able to check that their quotient does the job.