I found the dy/dx. But how do you find a vertical tangent line? And how do you find a vertical tangent when x = (?)?

\frac{dy}{dx}=\frac{2x-y}{x-3y^{2}}

What gives division by 0? When x=3y^{2}... [1]

Sub this back into the original function:

(3y^{2})^{2}-(3y^{2})y+y^{3}=10

9y^{4}-2y^{3}-10=0

Solving this for y gives:

y=-0.97533487561 and y=1.08709662767

Sub these into [1] gives:

x=2.85383435874 and x=3.5433723367

These are the coordinates of the points of vertical tangency.