Too many slopes eh?

We know that the slope of a curve is given by the derivative of the function at that point. So, $f'(x)=\dfrac{d}{dx}\dfrac{1}{(x-1)}=-\dfrac{1}{(x-1)^2}$

And here, we want the slope to be $-1$ . Therefore; $-\dfrac{1}{(x-1)^2}=-1\implies (x-1)^2=1\\x^2-2x=0\implies x(x-2)=0$

Hence $\boxed{x=0}$ or $\boxed{x=2}$ . Now all we need to do is find the value of $f(x)$ at these co-ordinates. $f(0) = \dfrac{1}{(0-1)} = \boxed{-1}\\ f(2) = \dfrac{1}{(2-1)} =\boxed 1$

Thus the co-ordinates of the points where the slope of the curve is $-1$ are: $(0,-1)$ and $(2,1)$ . So the sum is $0+(-1)+2+1=2$ .