\begin {align*} y^{\prime } & =\frac {-1}{1+x}y^{2}+\frac {1}{x-1}\\ y\left ( 0\right ) & =0 \end {align*}
\(f\left ( x,y\right ) =\frac {-1}{1+x}y^{2}+\frac {1}{x-1}\) is continuous in \(x\) everywhere except at \(x=-1\) and \(x=1\). And \(f_{y}=\frac {-2}{1+x}y\) is continuous except at \(x=-1\). Since
initial conditions at \(x_{0}=0,y_{0}=0\) then there is a unique solution in some rectangle inside the
rectangle \(-1