Pre-Image of a non continuous function. Let $f:\mathbb{R}\to\mathbb{R}$ be a function defined as follows:
$$f(x) = 
     \begin{cases}
       x^2 : x>0\\
       1-x : x\leq0\\
     \end{cases}$$
I have to calculate the pre-image of the interval $[-2,3)$
We know that the function's image are the positive real numbers, then we can simplify the interval this way: 
$$f^{-1}([-2,3))=f^{-1}((0,3))\cup f^{-1}([-2,0])=f^{-1}((0,3))$$
Because negative images don't exist.
To calculate $f^{-1}((0,3))$ I did this:
For the positive part of the function, $f(x)=x^2$
$f^{-1}((0,3))=(x^2=0,x^2=3)=(0,\sqrt3)$
For the negative part of the function, $f(x)=1-x$
$f^{-1}((0,3))=(1-x=0,1-x=3)=(1,-2)$
$f(x)=1-x$ is defined only for negative numbers and 0, then:
$f^{-1}((0,3))=(-2,1)-(0,1)=(-2,0]$
Finally, the pre-image of the function should be the sum of the two intervals:
$f^{-1}([-2,3)) = (0,\sqrt3)\cup (-2,0]=(-2,\sqrt3)$
But:
$f^{-1}(0) = a \longrightarrow f(a)=0 \longrightarrow
\begin{cases}
       0 = 1-a \rightarrow a=1 \\
       0 = a^2 \longrightarrow a=0\\
     \end{cases}$
$a=1$ is not a pre-image as $1>0$ then the interval $1-x$ of the function is not applied.
$a=0$ is not a pre-image as $f(0)=1$
From this I get that $f^{-1}(0)$ does not exist.
Then: $$f^{-1}([-2,3))=(-2,\sqrt3)-\{0\}$$

In the image it is shown the inverse of the function in the interval $-2\leq f(x)\leq 3$ which is in fact what I got, but the 0 is what I'm not sure about.

Yes you way is right but we need also to include the $x=0$ value in the preimage indeed
$$f(0)=1 \in [-2,3)$$
