Any non-constant morphism $\Phi\colon X\to Y$ between projective smooth curves has a degree $d$ and the morphism $\Phi$ will be an isomorphism if and only if that degree is $1$.
The incredibly good news is that you can calculate that degree by just looking at the fibre $\Phi^{-1}(y)$ of $\Phi$ at just one point ( any point!) of $y\in Y$: the degree is the dimension $d=dim_k \Gamma (\Phi^{-1}(y),\mathcal O)$ .
It is not terribly difficult to explain what the right-hand side means, but this is not even necessary here: we have $d=1$ as soon as for some non-empty $V\subset Y$ the restricted morphism $\Phi^{-1}(V)\to V$ is an isomorphism.
Since this is true in your case, we are done.
Edit
A more elementary proof (maybe the one Hartshorne had in mind at the level of Chapter 1, before "degree" is introduced) would be to consider the inverse isomorphism $\psi:V\to U$, to complete it to a morphism $\bar \psi:\mathbb P^1\to \mathbb P^1 $ (just as you did for $\phi$) and realize that $\bar \psi $ is an inverse to $\bar \phi $, which is thus an isomorphism.
Maybe it's time for a (reasonably complete) answer.
Consider the open subvariety $U_{YZ}:=\Bbb{P}^2 \setminus V(YZ)$ (i.e. $\Bbb{P}^2$ with both the $Y$-axis and the $Z$-axis removed). The rational map $f: \Bbb{P}^2 \dashrightarrow \Bbb{P}^2$ can be represented on $U_{YZ}$ by the morphism
\begin{split}
U_{YZ} &\rightarrow U_Z:=\Bbb{P}^2 \setminus V(Z)\\
(x:y:z) &\mapsto (x/z:x/y:1)
\end{split}
which shows that $f$ is defined on all of $U_{YZ}$. Similarly, you find that $f$ is defined on all of $\Bbb{P}^2 \setminus \{(1:0:0),(0:1:0),(0:0:1)\}$, and the only thing left to check for 1. and 2. is that $f$ cannot be extended to all of $\Bbb{P}^2$. One possible explanation is the following.
Consider the morphism
\begin{split}
f_{cone}: \Bbb{A}^3 &\rightarrow \Bbb{A}^3\\
(x,y,z) &\mapsto (xy,xz,yz)
\end{split}
which maps every point of the form $(\lambda,0,0),(0,\mu,0)$ or $(0,0,\nu)$ to the point $(0,0,0)$.
Like every morphism of varieties, $f_{cone}$ is already uniquely determined by its restriction to any non-empty open subset, e.g. by its restriction to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\} \subset \Bbb{A}^3$.
But $f_{cone}$ restricted to $\Bbb{A}^3 \setminus \{(\lambda,0,0),(0,\mu,0),(0,0,\nu) \; \vert \; \lambda,\mu,\nu \in k\}$ induces the morphism representing the rational map $f$, as described above. Hence every extension of $f$ to a morphism defined on all of $\Bbb{P}^2$ would have to map the points $(1:0:0),(0:1:0)$ and $(0:0:1)$ to "$(0:0:0)$", which is impossible, and consequently, the points $(1:0:0),(0:1:0),(0:0:1)$ cannot be in the domain of $f$.
For 3., you simply use the "trick" mentioned by Asal. Let $U_{XYZ}:=\Bbb{P}^2 \setminus V(XYZ)$ (i.e. $\Bbb{P}^2$ without the coordinate axes). On $U_{XYZ}$, we can represent $f$ as
$$
(x:y:z) \mapsto (xy:xz:yz)=1/(xyz)(xy:xz:yz)=(1/z:1/y:1/x)
$$
and from this, you can read off the inverse of $f_{\vert U_{XYZ}}$ directly.
Best Answer
Let $U = \mathbb{P}^2 \setminus \{P_0, P_1,P_2\}$ where $P_i$ are the points. We try to extend $\varphi$ past $U$. For points $p \in V_p(x_1)$ we have $$\varphi(p) = \varphi([p_0, 0, p_2]) = [0, p_0 p_2, 0] = [0,1,0].$$ For $p \in V_p(x_2)$ we see that the rational map gives $\varphi(p) = [0,0,1]$. Define $r_\lambda = [p, 0, \lambda]$ and $s_\lambda = [p, \lambda, 0]$. Then $$\lim_{\lambda \to 0} \varphi(r_\lambda) = [0,1,0] \text{ and } \lim_{\lambda \to 0} \varphi(s_\lambda) = [0,0,1].$$ But $\lim_{\lambda \to 0} r_\lambda = s_\lambda = [1,0,0]$ so that if $\varphi$ extends to $P_0$ then $\varphi(P_0) = [0,1,0] = [0,0,1]$ which is a contradiction. By symmetry we see that $\varphi$ does not extend to $P_1, P_2$.