Find the equations of all lines having slope $0$ which are tangents to the curve $y = \cfrac{1}{{{x^2} - 2x + 3}}.$

We have,
$y = \cfrac{1}{{{x^2} - 2x + 3}}$ …(i)

Differentiating (i), w.r.t. $x$, we get
$\cfrac{{dy}}{{dx}} = \cfrac{{ - 1}}{{{{({x^2} - 2x + 3)}^2}}}\cfrac{d}{{dx}}({x^2} - 2x + 3) = \cfrac{{ - (2x - 2)}}{{{{({x^2} - 2x + 3)}^2}}}$

For tangents having slope $0$, we must have $\cfrac{{dy}}{{dx}} = 0$
$\Rightarrow \cfrac{{ - (2x - 2)}}{{({x^2} - 2x + 3)}} = 0 \Rightarrow 2x - 2 = 0 \Rightarrow x = 1$

When $x = 1,y = \cfrac{1}{{{1^2} - 2 \cdot 1 + 3}} = \cfrac{1}{2}$ (using (i))

Therefore the tangent to the curve (i) at $\left( {1,\;\cfrac{1}{2}} \right)$ with slope $0$ will be given by

$y - \cfrac{1}{2} = 0(x - 1)$, or $2y - 1 = 0$, or $y = \cfrac{1}{2}$