Answer
$$ - \frac{1}{{\sqrt 3 }}{\text{ and }}\frac{1}{{\sqrt 3 }}$$
Work Step by Step
$$\eqalign{
& {x^2} + {y^2} = 1;\,\,\,\,\,\,\,\,\,\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right),\,\,\,\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right) \cr
& \left. {\text{a}} \right){\text{Solving for }}y \cr
& {y^2} = 1 - {x^2} \cr
& y = \pm \sqrt {1 - {x^2}} \cr
& \frac{{dy}}{{dx}} = \pm \frac{{ - 2x}}{{2\sqrt {1 - {x^2}} }} \cr
& \frac{{dy}}{{dx}} = \mp \frac{x}{{\sqrt {1 - {x^2}} }} \cr
& {\text{Find the slope at }}\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right){\text{ and }}\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right) \cr
& m = \mp \frac{{1/2}}{{\sqrt {1 - {{\left( {1/2} \right)}^2}} }} \cr
& m = \mp \frac{{1/2}}{{\sqrt {1 - {{\left( {1/2} \right)}^2}} }} \cr
& m = - \frac{1}{{\sqrt 3 }}{\text{ and }}\frac{1}{{\sqrt 3 }} \cr
& \cr
& \left. {\text{b}} \right){\text{By implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {{x^2} + {y^2}} \right] = \frac{d}{{dx}}\left[ 1 \right] \cr
& 2x + 2y\frac{{dy}}{{dx}} = 0 \cr
& \frac{{dy}}{{dx}} = - \frac{{2x}}{{2y}} \cr
& \frac{{dy}}{{dx}} = - \frac{x}{y} \cr
& {\text{Find the slope at }}\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right){\text{ and }}\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right) \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)}} = - \frac{{1/2}}{{\sqrt 3 /2}} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{1}{2},\frac{{\sqrt 3 }}{2}} \right)}} = - \frac{1}{{\sqrt 3 }} \cr
& and \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)}} = - \frac{{1/2}}{{ - \sqrt 3 /2}} \cr
& {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {\frac{1}{2}, - \frac{{\sqrt 3 }}{2}} \right)}} = \frac{1}{{\sqrt 3 }} \cr} $$
