Answer
$${m_1} = - 1{\text{ and }}{m_2} = 1$$
Work Step by Step
$$\eqalign{
& {y^2} + {x^2} = {y^4} - 2x \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {{y^2}} \right) + \frac{d}{{dx}}\left( {{x^2}} \right) = \frac{d}{{dx}}\left( {{y^4}} \right) - \frac{d}{{dx}}\left( {2x} \right) \cr
& {\text{find derivatives}} \cr
& 2y\frac{{dy}}{{dx}} + 2x = 4{y^3}\frac{{dy}}{{dx}} - 2 \cr
& {\text{solve for }}\frac{{dy}}{{dx}} \cr
& 2y\frac{{dy}}{{dx}} - 4{y^3}\frac{{dy}}{{dx}} = - 2x - 2 \cr
& \frac{{dy}}{{dx}} = \frac{{ - 2x - 2}}{{2y - 4{y^3}}} \cr
& \frac{{dy}}{{dx}} = \frac{{ - x - 1}}{{y - 2{y^3}}} \cr
& \cr
& {\text{Find the slope at the points }}\left( { - 2,1} \right){\text{ and }}\left( { - 2, - 1} \right) \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{\left( { - 2,1} \right)}} = \frac{{ - \left( { - 2} \right) - 1}}{{\left( 1 \right) - 2{{\left( 1 \right)}^3}}} = - 1 \cr
& {\left. {m = \frac{{dy}}{{dx}}} \right|_{\left( { - 2, - 1} \right)}} = \frac{{ - \left( { - 2} \right) - 1}}{{\left( { - 1} \right) - 2{{\left( { - 1} \right)}^3}}} = 1 \cr
& \cr
& m = - 1{\text{ and }}m = 1 \cr} $$
