Answer
$$\eqalign{
& {\bf{a}}){\text{ }}m = - 6 \cr
& {\bf{b}})y = - 6x - 11 \cr} $$
Work Step by Step
$$\eqalign{
& 3{x^2} - {y^2} = 11 \cr
& {\text{Differentiating each side w}}{\text{.r}}{\text{.t}} \hspace{2mm}x\cr
& 3\frac{d}{{dx}}\left[ {{x^2}} \right] - \frac{d}{{dx}}\left[ {{y^2}} \right] = \frac{d}{{dx}}\left[ {11} \right] \cr
& 3\left( {2x} \right) - 2y\frac{{dy}}{{dx}} = 0 \cr
& 6x - 2y\frac{{dy}}{{dx}} = 0 \cr
& - 2y\frac{{dy}}{{dx}} = - 6x \cr
& \frac{{dy}}{{dx}} = \frac{{ - 6x}}{{ - 2y}} \cr
& \frac{{dy}}{{dx}} = \frac{{3x}}{y} \cr
& \cr
& \left( {\bf{a}} \right){\text{ Calculating the slope at the given point}}\left( { - 2,1} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 2,1} \right)}} = \frac{{3\left( { - 2} \right)}}{1} \cr
& m = - 6 \cr
& \cr
& \left( {\bf{b}} \right) \cr
& {\text{Find the equation of the tangent line at }}\left( { - 2,1} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = - 6\left( {x + 2} \right) \cr
& y - 1 = - 6x - 12 \cr
& y = - 6x - 11 \cr} $$
