Let us look at a more familiar problem. We want to find a function $y$ such that $\frac{dy}{dt}=-\frac{1}{(t-3)^2} \qquad \text{and} \qquad y(4)=17$
Integrate. If we do it in the familiar first year calculus way, we get $1/(t-3)+C$. To evaluate $C$, put $t=4$. Quickly we get $y(t)=-1/(t-3)+18$.
For what $t$ is this correct? I claim that it is only for $t >3$. To see why, assume that $y$ represents the displacement of a particle at time $t$. Then $\frac{dy}{dt}$ represents its velocity.
The velocity (and displacement) are not defined at time $t=3$. From the information that the velocity at any time $t\ne 3$ is $-1/(t-3)^2$, and the information that the displacement at $t=4$ is $17$, we can find the displacement at any time $t>3$.
But we cannot go backwards past the singularity at time $t=3$ to reach any conclusion about displacement at any time $t<3$.
Similarly, if we had been told the displacement at say $t=1$, we could find the displacement at any time $t<3$, but we could get no information about $t>3$.
If we want a "general" formula for the $\int\frac{-dt}{(t-3)^2}$ that holds for as wide a domain as possible, we would have to say that $y=-1/(t-3)+C_1$ for $t>3$ and $y=-1/(t-3)+C_2$ for $t<3$, where $C_1$ and $C_2$ are constants, possibly different constants.
But more usually we restrict the domain, and like in our example with the initial condition $y(4)=17$, we simply say that $y(t)=-1/(t-3)+18$ for $t>3$.
Roughly speaking the same sort of thing happens in your differential equation example. You presumably solved the equation by rewriting as $\frac{dy}{y^2}=dx$ then integrating, and applying the initial condition.
Look at what you got for $y$. It (and its derivative) blows up at $x=1$. So like in the integration problem I started with, initial information about $x=0$ can't give any knowledge about the function past its singularity at $x=1$.
Suppose now we take your suggested $y(2)=-1$. When solving the DE, we get $-1/y=x+C$, so $C=-1$, I hope. That gives $y=-1/(x-1)$, valid for $x>1$.
But change things to $y(2)=1$. We get $C=-3$. So the position of the singularity can be affected by the initial condition. (This cannot happen in simple integration examples.)