The tangent to the curve $y^2 - xy + 9 = 0$ is vertical when

Given the equation of the curve:

$y^2 - xy + 9 = 0$

Differentiating both sides with respect to $x$, we get:

$2y \dfrac{dy}{dx} - \left( y + x \dfrac{dy}{dx} \right) = 0$

$\dfrac{dy}{dx} (2y - x) = y$

$\dfrac{dy}{dx} = \dfrac{y}{2y - x}$

For the tangent to be vertical, its slope $\dfrac{dy}{dx}$ must approach infinity. This occurs when the denominator is zero and the numerator is non-zero:

$2y - x = 0 \Rightarrow x = 2y$

Substituting $x = 2y$ into the equation of the curve:

$y^2 - (2y)y + 9 = 0$

$y^2 - 2y^2 + 9 = 0$

$-y^2 + 9 = 0$

$y^2 = 9$

$y = \pm 3$

For $y = \pm 3$, the numerator $y \neq 0$, so the tangent is indeed vertical.