 Solution

Work over the reals with $x\ne 0$, and put $p=dy/dx$. The algebraic equation implies $p\ne 0$ andDifferentiate this identity and use $y^{\prime }=p$; cancellation gives $dp/dx=p/x$. Hence $p=Cx$, with nonzero real $C$, and substitution gives the complete familyThere is no extra real singular branch lost in the cancellation, since $p^2+1\ne 0$. At a point $(x_0,y_0)$ with $x_0\ne 0$, the permitted constants satisfy $x_0^2{C}^2-2y_{0}C-1=0$, givingThese are distinct, real and nonzero, so exactly two integral curves pass through each such point. This does not contradict an Picard-Lindelöf theorem: the implicit equation permits two different slopes and is not one specified explicit first-order ordinary differential equation. Each choice of the smooth slope branch separately defines a locally unique curve.