Use implicit tangents and a normal to bound an area

Implicit differentiation gives $2yy^{\prime}-(y+xy^{\prime})=0$, hence $(2y-x)y^{\prime}=y$. A vertical tangent requires $2y-x=0$. Substituting $y=x/2$ in the curve gives $-x^2/4=-1$, so $x=\pm2$. At $(2,1)$ and $(-2,-1)$ the numerator $y$ is nonzero, so the tangent lines are $x=2$ and $x=-2$.

A zero derivative would require $y=0$, but this makes $y^2-xy=0$, contradicting $-1$. The curve therefore has no stationary points.

When $y=2$, $4-2x=-1$ gives $x=5/2$. The tangent gradient is $2/(4-5/2)=4/3$, so the normal gradient is $-3/4$. Its equation is

$y-2=-\frac34\left(x-\frac52\right),\qquad y=-\frac34x+\frac{31}{8}.$

The intercepts are $(0,31/8)$ and $(31/6,0)$. The bounded region is a right triangle of area $\tfrac12(31/6)(31/8)=961/96$ square units.