Find the point of intersection of the tangents to the curve y^2 -3xy + x^3 = 3 at the points where x=-1.

Hello!
The steps are: 1) find the tangent points, 2) find the equations of the tangents, 3) find the intersection of the tangents.
1. It is known that x=-1. Substitute it to the curve equation and obtain an equation for y:
y^2+3y-1=3, or y^2+3y-4=0.
The solutions are y_1=1 and y_2=-4. So the points of tangent are T_1(-1,1) and T_2(-1,-4).
2. Differentiate the equation d/(dx):
2yy' - 3y - 3xy' + 3x^2 = 0, or y'(2y - 3x) = 3y - 3x^2, or
y'=(3(y-x^2))/(2y-3x).
For T_1 we obtain that y'=0. Thus the tangent is a horizontal line with the equation y=1.
For T_2 we obtain y'=(3(-4 - 1))/(2*(-4) - 3*(-1)) = -15/(-5) = 3. Thus the equation is y=3(x+1)-4 = 3x-1.
3. The point of intersection has y=1 and x from the equation 1=3x-1, i.e. x=2/3. So the answer is the point (2/3, 1).
The graph is at the link.
https://www.desmos.com/calculator/touovjs2zr