Answer
$$\frac{{dy}}{{dx}} = \frac{{{x^2} - {y^2}}}{{2xy - {y^2}}}$$
Work Step by Step
$$\eqalign{
& {x^3} - 3x{y^2} + {y^3} = 5 \cr
& \cr
& {\text{Let }}f\left( {x,y} \right) = {x^3} - 3x{y^2} + {y^3} \cr
& {\text{calculate the partial derivatives }}\frac{{\partial f}}{{\partial x}}{\text{ and }}\frac{{\partial f}}{{\partial y}} \cr
& \frac{{\partial f}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {{x^3} - 3x{y^2} + {y^3}} \right] \cr
& {\text{treat }}y{\text{ as a constant}} \cr
& \frac{{\partial f}}{{\partial x}} = 3{x^2} - 3{y^2} \cr
& and \cr
& \frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {{x^3} - 3x{y^2} + {y^3}} \right] \cr
& {\text{treat }}x{\text{ as a constant}} \cr
& \frac{{\partial f}}{{\partial y}} = - 6xy + 3{y^2} \cr
& \cr
& {\text{Use the theorem 13}}{\text{.5}}{\text{.3 to find }}dy/dx \cr
& \frac{{dy}}{{dx}} = - \frac{{\partial f/\partial x}}{{\partial f/\partial y}} \cr
& \frac{{dy}}{{dx}} = - \frac{{3{x^2} - 3{y^2}}}{{ - 6xy + 3{y^2}}} \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{{{x^2} - {y^2}}}{{2xy - {y^2}}} \cr} $$
