Answer
$$y = - \frac{5}{9}x + \frac{{16}}{9}$$
Work Step by Step
$$\eqalign{
& y = \frac{x}{{{x^2} - 1}},\,\,\,\,\,x = 2 \cr
& {\text{find the derivative of }}y \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{x}{{{x^2} - 1}}} \right] \cr
& {\text{by using the quotient rule}} \cr
& \frac{{dy}}{{dx}} = \frac{{\left( {{x^2} - 1} \right)\left( 1 \right) - x\left( {2x} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2} - 1 - 2{x^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - 1 - {x^2}}}{{{{\left( {{x^2} - 1} \right)}^2}}} \cr
& {\text{Find the slope of the tangent line at }}x = 2 \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{x = 2}} \cr
& m = \frac{{ - 1 - {{\left( 2 \right)}^2}}}{{{{\left( {{2^2} - 1} \right)}^2}}} \cr
& m = \frac{{ - 5}}{9} \cr
& \cr
& {\text{Evaluate the function at }}x = 2 \cr
& y\left( 2 \right) = \frac{2}{{{2^2} - 1}} \cr
& y\left( 2 \right) = \frac{2}{3} \cr
& {\text{we know the point }}\left( {2,\frac{2}{3}} \right){\text{ and the slope }}m = - \frac{5}{9} \cr
& {\text{find the equation of the tangent line using the point - slope form of a line}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - \frac{2}{3} = - \frac{5}{9}\left( {x - 2} \right) \cr
& {\text{simplifying}} \cr
& y - \frac{2}{3} = - \frac{5}{9}x + \frac{{10}}{9} \cr
& y = - \frac{5}{9}x + \frac{{10}}{9} + \frac{2}{3} \cr
& y = - \frac{5}{9}x + \frac{{16}}{9} \cr} $$
