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?
 A: $\newcommand{\P}{\Bbb P}$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 ! 
There is a very nice interpretation in term of the Veronese embedding and linear systems of conics. 
