Here is the solution for this specific problem:
Based from the graph, the curve will intersect itself at the y-axis, i.e. x = 0.
t^3 - 6t = 0
t(t^2 - 6) = 0
t = 0 or t = ± √6
dx/dt = 3t^2 - 6
dy/dt = 2t
dy/dx = 2t/(3t^2 - 6)
@ t = 0, dy/dx = 0.
x = 0, y = 0
y = 0
@ t = √6, dy/dx = 2√6/12 = √6/6
x = 0, y = 6
y - 6 = (√6/6) x
y = (√6/6)x + 6
@ t = -√6, dy/dx = -2√6/12 = -√6/6
x = 0, y = 6
y - 6 = (-√6/6) x
y = (-√6/6)x + 6
So the equations of the tangent line at the point where the curve crosses itself are: y = (√6/6)x + 6 and y = (-√6/6)x + 6. I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.