Answer
$$\,y = - \frac{{24}}{{43}}x + \frac{{225}}{{43}}$$
Work Step by Step
$$\eqalign{
& {x^2}y + {y^3} = 75{\text{ point }}\left( {4,3} \right) \cr
& {\text{use implicit differentiation}}{\text{, differentiate both sides with respect to }}x \cr
& \frac{d}{{dx}}\left[ {{x^2}y} \right] + \frac{d}{{dx}}\left[ {{y^3}} \right] = \frac{d}{{dx}}\left[ {75} \right] \cr
& {\text{using the product rule}} \cr
& {x^2}\frac{d}{{dx}}\left[ y \right] + y\frac{d}{{dx}}\left[ {{x^2}} \right] + \frac{d}{{dx}}\left[ {{y^3}} \right] = \frac{d}{{dx}}\left[ {75} \right] \cr
& {\text{solve derivatives}} \cr
& {x^2}y'\left( x \right) + y\left( {2x} \right) + 3{y^2}y'\left( x \right) = 0 \cr
& {\text{simplify}} \cr
& {x^2}y'\left( x \right) + y\left( {2x} \right) + 3{y^2}y'\left( x \right) = 0 \cr
& {\text{solve for }}y'\left( x \right) \cr
& {x^2}y'\left( x \right) + 2xy + 3{y^2}y'\left( x \right) = 0 \cr
& {x^2}y'\left( x \right) + 3{y^2}y'\left( x \right) = - 2xy \cr
& \left( {{x^2} + 3{y^2}} \right)y'\left( x \right) = - 2xy \cr
& y'\left( x \right) = - \frac{{2xy}}{{{x^2} + 3{y^2}}} \cr
& {\text{find the slope at the point }}\left( {4,3} \right) \cr
& \,\,\,m = - \frac{{2\left( 4 \right)\left( 3 \right)}}{{{{\left( 4 \right)}^2} + 3{{\left( 3 \right)}^2}}} \cr
& \,\,\,m = - \frac{{24}}{{43}} \cr
& {\text{find the equation of the tangent line at the point }}\left( {4,3} \right) \cr
& \,\,\,\,\,y - {y_1} = m\left( {x - {x_1}} \right) \cr
& \,\,\,\,\,y - 3 = - \frac{{24}}{{43}}\left( {x - 4} \right) \cr
& {\text{simplify}} \cr
& \,\,\,\,\,y - 3 = - \frac{{24}}{{43}}x + \frac{{96}}{{43}} \cr
& \,\,\,\,\,y = - \frac{{24}}{{43}}x + \frac{{96}}{{43}} + 3 \cr
& \,\,\,\,\,y = - \frac{{24}}{{43}}x + \frac{{225}}{{43}} \cr} $$
