Prove that $\phi$ is a birational map and that the inverse map coincides with the blowup.

This is Exercise II.4.4 in Shafarevich's book Basic Algebraic Geometry, second edition.

Consider the rational map $\phi:\mathbb{P}^2\to \mathbb{P}^4$ given by $$\phi(x_0:x_1:x_2)=(x_0x_1:x_0x_2:x_1^2:x_1x_2:x_2^2).$$ Prove that $\phi$ is a birational map to a surface $\overline{\phi(\mathbb{P}^2)}$, and that the inverse map $\overline{\phi(\mathbb{P}^2)}\to \mathbb{P}^2$ coincides with the blowup.

Here is what I have done. I called the variables in $\mathbb{P}^2$ $z_0, \cdots, z_4$ and obtained the equations \begin{align*} z_0&=x_0x_1\\ z_1&=x_0x_2\\ z_2&=x_1^2\\ z_3&=x_1x_2\\ z_4&=x_2^2. \end{align*} Then I have tried to solve this for $x_0, x_1$ and $x_2$ to obtain a map $\psi:\overline{\phi(\mathbb{P}^2)}\to \mathbb{P}^2$. However, it doesn't seem to be possible to do this, I always get expressions for $x_i^2$, not $x_i$. Of course, I can't take roots. Then I tried to make some assumptions. Since $\phi$ is not defined for $p=(1:0:0)$, I assume that $x_1=1$ and I obtain $$\psi(z_0:z_1:z_2:z_3:z_4)=(z_0:1:z_3).$$ By assuming that $x_2=1$ instead, I get a similar (but different) map. In both cases, $\psi$ is not rational and they "use" only two of the five variables. So I doubt that what I have done is the right way to go. So, now I am stuck. Can someone please help me?

$\newcommand{\P}{\Bbb P}$Outside $[1:0:0]$, $\phi$ is an isomorphism. For example, an partial inverse is given by (with your notation) $(z_0:z_1:z_2:z_3:z_4) \mapsto (z_0:z_1:z_3)$. This works only if $x_1 \neq 0$ (or equivalently $z_0,z_1,z_3$ are not all zero). Other cases are similar.
So we just need to understand what is the closure of the image. Since we understand the image of $v_2(\P^2 \backslash \{p\})$ we just need to see what is $\lim\limits_{x \to p} \phi(x)$. For this, I take the path $x$ defined by $$x_0(t) = 1, x_1(t) = at, x_2(t) = bt$$ for some $a,b \in k$. It is easy to get that $\lim\limits_{t \to 0}\phi(x(t)) = [a:b:0:0:0]$. This shows that $\phi$ is remplacing a point by a projective line : this is a good start for the blow-up !
For verify it we just need to understand $\phi$ around $[1:0:0]$. Recall that the local form of the blow-up is the map $\Bbb A^2 \to \Bbb A^2 \times \P^1$ $(x,y) \to ((x,y), [x:y])$. Back to your map, take local coordinates around $[1:0:0]$, namely $s=\frac{x_1}{x_0}$ and $t = \frac{x_2}{x_0}$. The map is just $(s,t) \mapsto [s:t:s^2:st:t^2]$. Now the three last coordinates corresponds to $(x,y)$ in our local model (i.e exercise : $\Bbb A^2 \backslash \{(0,0)\} \to \P^2, (x,y) \mapsto (x^2;xy:y^2)$ is an embedding) and of course the first part is remembering the direction of $(s,t)$. So this is exactly the blow-up !