Step 1:
$y=\large\frac{x-1}{x-2}$
Differentiating w.r.t $x$ on both sides,by applying the quotient rule,
$\large\frac{dy}{dx}=\frac{(x-2)(1)-(x-1)(1)}{(x-2)^2}$
$\qquad=\large\frac{x-2-x+1}{(x-2)^2}$
$\qquad=\large\frac{-1}{(x-2)^2}$
Step 2:
$\large\frac{dy}{dx}$ at $x=10$
Substituting for $x$ we get,
$\large\frac{dy}{dx}_{(x=10)}=\large\frac{-1}{(10-2)^2}$
$\qquad\qquad=\large\frac{-1}{64}$
Hence the slope of the tangent to the curve $y=\large\frac{x-1}{x-2}$ is $\large\frac{-1}{64}$