Answer
$$\frac{{dy}}{{dx}} = \frac{1}{{2y - 1}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{x + 1}}{{y - 1}} \cr
& \left( {y - 1} \right)y = x + 1 \cr
& {\text{Use the implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\left( {y - 1} \right)y} \right] = \frac{d}{{dx}}\left[ {x + 1} \right] \cr
& \left( {y - 1} \right)\frac{d}{{dx}}\left[ y \right] + y\frac{d}{{dx}}\left[ {y - 1} \right] = \frac{d}{{dx}}\left[ {x + 1} \right] \cr
& \left( {y - 1} \right)\frac{{dy}}{{dx}} + y\frac{{dy}}{{dx}} = 1 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& \left[ {\left( {y - 1} \right) + y} \right]\frac{{dy}}{{dx}} = 1 \cr
& \left( {2y - 1} \right)\frac{{dy}}{{dx}} = 1 \cr
& \frac{{dy}}{{dx}} = \frac{1}{{2y - 1}} \cr} $$
