Answer
$${\text{ }}{\bf{a}}){\text{ }}m = \frac{3}{{10}},{\text{ }}{\bf{b}})y = \frac{3}{{10}}x - 4$$
Work Step by Step
$$\eqalign{
& {x^2} - 10xy = 200 \cr
& {\text{We have that }}x = 10,{\text{ substituting}} \cr
& {\left( {10} \right)^2} - 10\left( {10} \right)y = 200 \cr
& {\text{Solve for }}y \cr
& 100 - 100y = 200 \cr
& y = - 1 \cr
& {\text{For }}x = 10{\text{ we obtain the point }}\left( {10, - 1} \right) \cr
& \cr
& {x^2} - 10xy = 200 \cr
& {\text{Differentiating each side w}}{\text{.r}}{\text{.t}} \hspace{2mm} x \cr
& \frac{d}{{dx}}\left[ {{x^2}} \right] - \underbrace {\frac{d}{{dx}}\left[ {10xy} \right]}_{{\text{Product rule}}} = \frac{d}{{dx}}\left[ {200} \right] \cr
& 2x - 10x\frac{{dy}}{{dx}} - 10y = 0 \cr
& {\text{Solve for }}\frac{{dy}}{{dx}} \cr
& - 10x\frac{{dy}}{{dx}} = - 2x + 10y \cr
& \frac{{dy}}{{dx}} = \frac{{2x - 10y}}{{10x}} \cr
& \frac{{dy}}{{dx}} = \frac{{x - 5y}}{{5x}} \cr
& \cr
& \left( {\bf{a}} \right){\text{ Calculating the slope at the point}}\left( {10, - 1} \right) \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( {10, - 1} \right)}} = \frac{{10 - 5\left( { - 1} \right)}}{{5\left( {10} \right)}} \cr
& m = \frac{3}{{10}} \cr
& \left( {\bf{b}} \right){\text{The equation of the tangent line at }}\left( {10, - 1} \right){\text{ is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y + 1 = \frac{3}{{10}}\left( {x - 10} \right) \cr
& y + 1 = \frac{3}{{10}}x - 3 \cr
& y = \frac{3}{{10}}x - 4 \cr} $$
