Find the slope of a normal line to a curve given curve equation and x coordinate

Question

What are the steps to solving this problem? (Thank you in advance)

Find the slope of the normal line to the curve (x^2)-(xy)+(y^2) = 7 at the point where x=1. There are two possible answers – accept the SMALLER answer.

Where I started:

y' = (y-2x) / (2y-x)

Answer 1

Slope of the normal at a point $P =m= -\frac{dx}{dy}|_P = -\frac{1}{\frac{dy}{dx}}\bigg|_P = -\frac{1}{y'|_P}$

As $x^2 - xy + y^2 =7$, by differentiating w.r.t $x$,

$2x - xy' - y + 2yy' = 0$

$y' = \frac{2x-y}{x-2y}$

and $$-\frac{1}{y'} = -\frac{x-2y}{2x-y}$$

At $x=1$, $1 -y+y^2 = 7 \implies y^2 -y -6 = 0\implies y = 3,-2$

At $(1,3)$ the slope is,

$m_1 = -\frac{1}{y'} = -\frac{1-6}{2-3} = -5$

At $(1,-2)$,

$m_2 = -\frac{1}{y'} = -\frac{1+4}{2+2} = -\frac{5}{4} $