The solution is not only discontinuous at $x=-2$ and $x=3$, it is also undefined at these points. The maximal interval containing $0$ on which there exists a solution is the open interval $(-2,3)$.
Edit to answer a comment: More generally the ODE around a point $x$ such that $y(x)\ne0$ is equivalent to y'/y^2=1-2x, that is, (1/y)'=(x^2-x)'. Hence, on every interval where $y$ is defined and not identically zero, there exists a constant $c$ such that $y(x)=1/(x^2-x+c)$. The value of $c$ depends on the initial condition one is given, hence the interval on which $y$ is defined also depends on the initial condition.
In your example, $y(0)=-1/6$, hence $c=-6$, $y(x)=1/(x^2-x-6)$ and the maximal interval of definition around $0$ is bounded by the roots of $x^2-x-6=(x+2)(x-3)$ which are closest to $0$, namely $-2$ and $3$.
But consider another example: if $y(-3)=-1/8$, then $c=-20$, $y(x)=1/(x^2-x-20)$ and the maximal interval of definition around $-3$ is bounded by the roots of $x^2-x-20=(x-5)(x+4)$ which are closest to $-3$, hence this interval is $(-4,+5)$.
The interval can also be semi-infinite or infinite. For instance, if $y(-3)=1/15$, then $c=3$, $y(x)=1/(x^2-x+3)$ and the maximal interval of definition around $-3$ is the real line $(-\infty,+\infty)$. If $y(-3)=1/12$, then $c=0$, $y(x)=1/(x^2-x)$ and the maximal interval of definition around $-3$ is the half line $(-\infty,0)$.
And so on.