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
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.