Solutions to ((\[DifferentialD]y)/(\[DifferentialD]x))=((x^2)/(1-(y^2))), are functional subsets of the equation y(x)=-((x^3)/2)+y-((y^3)/3).  When given an initial value, the largest domain where the piece of the level curve containing that point is a function is considered the particular solution to the IVP.

plot [K_]:=ContourPlot[-x^3/3+y-y^3/3==K,{x,-3,3},{y,-3,3},Axes->True,AxesOrigin->{0,0},Frame->False,ContourStyle->{Thick,Blue}]
rpoint[x_,y_]:= Graphics[{Red,PointSize[Large],Point[{x,y}]}];
tangent[x_,y_]:= ParametricPlot[{x,y+t},{t,-.5,.5},PlotStyle->{Orange,Thick}]
domain = Plot[0,{x,-2^(1/3),2^(1/3)},PlotStyle->{Red,Thickness[.02]}];
Show[plot[0],plot[1],plot[-1],plot[.5],plot[-.5]]

Here, the particular solution to the IVP ((\[DifferentialD]y)/(\[DifferentialD]x))=((x^2)/(1-(y^2))), x ∈ (-(2^(1/3)),(2^(1/3))). It is understood here that the initial value tells you the proper branch of the curve, as can be seen here as between the places where the level curve has a vertical tangent.

Show[plot[0],plot[1],plot[-1],plot[.5],plot[-.5],rpoint[2^(1/3),1],rpoint[-2^(1/3),-1],tangent[2^(1/3),1],tangent[-2^(1/3),-1], domain]