First, let's put in a few values of x and y: x=0: y=-1; y=0, x=1; x=2, 4y+2y^2+2-y-1=0, 2y^2+3y+1=0=(2y+1)(y+1), y=-1/2 and -1; x=-1: -y^2+y-1-y-1=0, -y^2-2=0, no real value for y. This gives us the points (0,-1), (1,0), (2,-1/2), (2,-1). Note that there are two values of x when y=-1, two values of y when x=2.
Treating the function as a quadratic in y: xy^2+y(x^2-1)+x-1=0, using the formula:
y=(1-x^2+sqrt((1-x^2)^2-4x(x-1)))/(2x)=(1-x^2+sqrt(x^4-6x^2+4x+1))/(2x). When x=0 y=-1, but this formula otherwise covers all x<>0. It is a very unusual graph. Note that y is defined only when the square root is positive, that is, when the quartic is positive: x^4-6x^2+4x+1>0. This factorises: (x-1)(x^3+x^2-5x-1)>0.
Differentiating the function: 2xy+x^2dy/dx+2xydy/dx+y^2+1-dy/dx=0, so dy/dx=-(y^2+2xy+1)/(x^2+2xy-1).
So we can calculate dy/dx at each point:
(0,-1): -2/-1=2
(1,0): -1/0
(2,-1/2): -(1/4-2+1)/(4-2-1)=3/4
(2,-1): -(1-4+1)/(4-4-1)=2/-1=-2
These are the slopes of the tangents at each of the 4 points. (1,0) is indeterminate suggesting a vertical tangent, which implies a vertex as in a parabola lying on its side. The slopes help to give some idea of the shape of the curve at various points. Notice where a negative slope is near to a positive slope, because there will be a turning point in between.
dy/dx=0 at a turning point, and the condition for this is y^2+2xy+1=0, or x=-(y^2+1)/2y, bearing in mind that some values of x have no real y value. Also, y=(-2x+sqrt(4x^2-4))/2=-x+sqrt(x^2-1).
The points above can be plotted with their tangents.