Answer
$$y = \frac{{{{\left( {{x^3} - 1} \right)}^3}}}{9} + 1$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {x^2}{\left( {{x^3} - 1} \right)^2} \cr
& {\text{Separate the variables}} \cr
& dy = {x^2}{\left( {{x^3} - 1} \right)^2}dx \cr
& {\text{Integrate both sides with respect to }}x \cr
& \int {dy} = \int {{x^2}{{\left( {{x^3} - 1} \right)}^2}} dx \cr
& \int {dy} = \frac{1}{3}\int {{{\left( {{x^3} - 1} \right)}^2}\left( {3{x^2}} \right)} dx \cr
& {\text{Let }}u = 4 - {x^2},{\text{ }}du = - 2xdx \cr
& \int {dy} = \frac{1}{3}\int {\overbrace {{{\left( {{x^3} - 1} \right)}^2}}^{{u^2}}\overbrace {\left( {3{x^2}} \right)dx}^{du}} \cr
& y = \frac{1}{3}\left[ {\frac{{{{\left( {{x^3} - 1} \right)}^3}}}{3}} \right] + C \cr
& y = \frac{{{{\left( {{x^3} - 1} \right)}^3}}}{9} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Use the initial condition }}\left( {1,0} \right) \cr
& 0 = \frac{{{{\left( {{1^3} - 1} \right)}^3}}}{9} + C \cr
& 1 = 0 + C \cr
& C = 1 \cr
& {\text{Substitute 2 for }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& y = \frac{{{{\left( {{x^3} - 1} \right)}^3}}}{9} + 1{\text{ }}\left( {{\text{Particular solution}}} \right) \cr} $$
