Answer
$$y = \frac{1}{{12}}x + \frac{5}{4}$$
Work Step by Step
$$\eqalign{
& 8{y^3} - 4x{y^2} = 20{\text{ at the point }}\left( { - 3,1} \right) \cr
& {\text{take the derivative on both sides with respect to }}x \cr
& \frac{d}{{dx}}\left( {8{y^3} - 4x{y^2}} \right) = \frac{d}{{dx}}\left( {20} \right) \cr
& \frac{d}{{dx}}\left( {8{y^3}} \right) - \frac{d}{{dx}}\left( {4x{y^2}} \right) = \frac{d}{{dx}}\left( {20} \right) \cr
& {\text{use product rule}} \cr
& \frac{d}{{dx}}\left( {8{y^3}} \right) - 4x\frac{d}{{dx}}\left( {{y^2}} \right) - {y^2}\frac{d}{{dx}}\left( {4x} \right) = \frac{d}{{dx}}\left( {20} \right) \cr
& {\text{solve the derivatives }} \cr
& 24{y^2}\frac{{dy}}{{dx}} - 4x\left( {2y} \right)\frac{{dy}}{{dx}} - {y^2}\left( 4 \right) = 0 \cr
& 24{y^2}\frac{{dy}}{{dx}} - 8xy\frac{{dy}}{{dx}} - 4{y^2} = 0 \cr
& {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr
& \left( {24{y^2}\frac{{dy}}{{dx}} - 8xy\frac{{dy}}{{dx}}} \right) = 4{y^2} \cr
& \left( {24{y^2} - 8xy} \right)\frac{{dy}}{{dx}} = 4{y^2} \cr
& \frac{{dy}}{{dx}} = \frac{{4{y^2}}}{{24{y^2} - 8xy}} \cr
& {\text{find the slope at the given point }} \cr
& m = {\left. {\frac{{dy}}{{dx}}} \right|_{\left( { - 3,1} \right)}} = \frac{{4{{\left( 1 \right)}^2}}}{{24{{\left( 1 \right)}^2} - 8\left( { - 3} \right)\left( 1 \right)}} = \frac{1}{{12}} \cr
& {\text{find the equation of the tangent line at the point }}\left( { - 3,1} \right) \cr
& {\text{use the point - slope form }}y - {y_1} = m\left( {x - {x_1}} \right) \cr
& y - 1 = \frac{1}{{12}}\left( {x + 3} \right) \cr
& y - 1 = \frac{1}{{12}}x + \frac{1}{4} \cr
& y = \frac{1}{{12}}x + \frac{5}{4} \cr} $$
