Answer
$$\left( {\text{a}} \right)dy = - \frac{{3{x^2}}}{{{{\left( {{x^3} - 1} \right)}^2}}}dx{\text{ and }}\left( {\text{b}} \right)\,dy = \frac{{ - 6{x^2} + 2{x^3} + 1}}{{{{\left( {2 - x} \right)}^2}}}dx$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)y = \frac{1}{{{x^3} - 1}} \cr
& \,\,\,\,\,y = {\left( {{x^3} - 1} \right)^{ - 1}} \cr
& \,{\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = - {\left( {{x^3} - 1} \right)^{ - 2}}\left( {3{x^2}} \right) \cr
& \,\,\,\frac{{dy}}{{dx}} = - \frac{{3{x^2}}}{{{{\left( {{x^3} - 1} \right)}^2}}} \cr
& {\text{Then}} \cr
& \,\,dy = - \frac{{3{x^2}}}{{{{\left( {{x^3} - 1} \right)}^2}}}dx \cr
& \cr
& \left( {\text{b}} \right)y = \frac{{1 - {x^3}}}{{2 - x}} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \,\,\,\frac{{dy}}{{dx}} = \frac{{\left( {2 - x} \right)\left( { - 3{x^2}} \right) - \left( {1 - {x^3}} \right)\left( { - 1} \right)}}{{{{\left( {2 - x} \right)}^2}}} \cr
& {\text{Then}} \cr
& \,\,\,\frac{{dy}}{{dx}} = \frac{{ - 6{x^2} + 3{x^3} + 1 - {x^3}}}{{{{\left( {2 - x} \right)}^2}}} \cr
& \,\,\,dy = \frac{{ - 6{x^2} + 2{x^3} + 1}}{{{{\left( {2 - x} \right)}^2}}}dx \cr
& \cr
& \left( {\text{a}} \right)dy = - \frac{{3{x^2}}}{{{{\left( {{x^3} - 1} \right)}^2}}}dx{\text{ and }}\left( {\text{b}} \right)\,dy = \frac{{ - 6{x^2} + 2{x^3} + 1}}{{{{\left( {2 - x} \right)}^2}}}dx \cr} $$
