Consider the curve given by the equation $y^3 - xy = 2$. It can be shown that $\dfrac{dy}{dx} = \dfrac{y}{3y^2 - x}$.
(a) Write an equation for the line tangent to the curve at the point $(-1, 1)$.
(b) Find the coordinates of all points on the curve at which the line tangent to the curve at that point is vertical.
(c) Evaluate $\dfrac{d^2y}{dx^2}$ at the point on the curve where $x = -1$ and $y = 1$.