Answer
$$y = - \frac{3}{5}x - 10$$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{64}} = 1;{\text{ }}\left( { - 6, - \frac{{32}}{5}} \right) \cr
& {\text{Find }}\frac{{dy}}{{dx}}{\text{ using the implicit differentiation}} \cr
& \frac{d}{{dx}}\left[ {\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{64}}} \right] = \frac{d}{{dx}}\left[ 1 \right] \cr
& \frac{{2x}}{{100}} + \frac{{2y}}{{64}}\frac{{dy}}{{dx}} = 0 \cr
& \frac{x}{{50}} + \frac{y}{{32}}\frac{{dy}}{{dx}} = 0 \cr
& \frac{y}{{32}}\frac{{dy}}{{dx}} = - \frac{x}{{50}} \cr
& \frac{{dy}}{{dx}} = - \frac{{16x}}{{25y}} \cr
& {\text{Calculate the slope at the point }}\left( { - 6, - \frac{{32}}{5}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{{16\left( { - 6} \right)}}{{25\left( { - 32/5} \right)}} \cr
& m = - \frac{3}{5} \cr
& {\text{The equation of the tangent line at the given point is}} \cr
& y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y + \frac{{32}}{5} = - \frac{3}{5}\left( {x + 6} \right) \cr
& y + \frac{{32}}{5} = - \frac{3}{5}x - \frac{{18}}{5} \cr
& y = - \frac{3}{5}x - 10 \cr
& \cr
& {\text{Graph}} \cr} $$
