AP calculus free response question

Consider the curve given by x^2+ 3y^2=1+3xy.

A) Show that dy/dx=(3y-2x)/(6y-3x)

B) Find all points on the curve whose x coordinate is 1, and write an equation for the tangent line at each of these points.

C) Find the coordinates of each point on the curve where the tangent line is vertical.

Thank you

(A) By differentiating both sides with respect to x, we get:
(d/dx)(x^2 + 3y^2) = (d/dx)(1 + 3xy)
==> 2x + 6y(dy/dx) = 3y + 3x(dy/dx), by the Chain and Product Rules
==> 6y(dy/dx) - 3x(dy/dx) = 3y - 2x, by bringing all terms involving dy/dx to one side
==> (dy/dx)(6y - 3x) = 3y - 2x, by factoring out dy/dx
==> dy/dx = (3y - 2x)/(6y - 3x), as required.

(B) Plugging in x = 1 into the curve and solving for y gives:
1 + 3y^2 = 1 + 3y ==> y = 0 and y = 1.

So, we have the points (1, 0) and (1, 1).

Plugging these points into dy/dx above gives the slope of the tangent line at each to be:
(a) (1, 0) ==> dy/dx = (0 - 2)/(0 - 3) = 2/3
(b) (1, 1) ==> dy/dx = (3 - 2)/(6 - 3) = 1/3.

Using point-slope form, the equations of the tangent lines are:
y = (2/3)(x - 1) = (2/3)x - 2/3 and y = (1/3)(x - 1) + 1 = (1/3)x + 2/3.

(C) The tangent line is vertical when the denominator of dy/dx is zero; in other words, the tangent line is vertical when:
6y - 3x = 0 ==> x = 2y.

Substituting x = 2y into the curve gives:
(2y)^2 + 3y^2 = 1 + 3(2y)y
==> 7y^2 = 1 + 6y^2.

This equation solves to give y = ±1. Then, since x = 2y, x = ±2.

Therefore, the required points here are (2, 1) and (-2, -1).

I hope this helps!