Answer
$$\eqalign{
& \left( {\bf{a}} \right) \cr
& y = x - 1 \cr
& y = - x + 2 \cr
& \left( {\bf{b}} \right){\text{ See the graph}} \cr} $$
Work Step by Step
$$\eqalign{
& x + {y^2} - y = 1;\,\,\,x = 1 \cr
& {\text{Calculate }}y{\text{ for }}x = 1 \cr
& 1 + {y^2} - y = 1 \cr
& {\text{ }}{y^2} - y = 0 \cr
& y\left( {y - 1} \right) = 0 \cr
& y = 0,{\text{ }}y = 1 \cr
& {\text{We obtain the points}} \cr
& \left( {1,0} \right),{\text{ }}\left( {1,1} \right) \cr
& {\text{Differentiate implicitly with respect to }}x \cr
& 1 + 2y\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = 0 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& 2y\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = - 1 \cr
& \frac{{dy}}{{dx}}\left( {2y - 1} \right) = - 1 \cr
& \frac{{dy}}{{dx}} = \frac{1}{{1 - 2y}} \cr
& {\text{Calculate the slope }}m{\text{ for the points }}\left( {1,0} \right),{\text{ }}\left( {1,1} \right) \cr
& \left( {1,0} \right) \Rightarrow m = \frac{1}{{1 - 2\left( 0 \right)}} = 1 \cr
& \left( {1,1} \right) \Rightarrow m = \frac{1}{{1 - 2\left( 1 \right)}} = - 1 \cr
& {\text{Find the equations of the tangent lines at the points }} \cr
& \left( {1,0} \right),{\text{ }}\left( {1,1} \right),{\text{ then}} \cr
& {\bf{For }}\left( {{\bf{1,0}}} \right) \cr
& y - 0 = 1\left( {x - 1} \right) \cr
& y = x - 1 \cr
& {\bf{For }}\left( {{\bf{1,1}}} \right) \cr
& y - 1 = - 1\left( {x - 1} \right) \cr
& {\text{ }}y = - x + 2 \cr} $$
