Answer
$$\frac{1}{{{y^2}}} = \frac{1}{3} + C{e^{2{x^3}}}$$
Work Step by Step
$$\eqalign{
& y' + 3{x^2}y = {x^2}{y^3} \cr
& {\text{The differential equation is written in the form}} \cr
& y' + P\left( x \right)y = Q\left( x \right){y^n} \cr
& y' + 3{x^2}y = {x^2}{y^3} \to P\left( x \right) = 3{x^2},{\text{ }}Q\left( x \right) = {x^2},{\text{ }}n = 3 \cr
& {\text{From the book we have that the general solution of a Bernoulli }} \cr
& {\text{equation is:}} \cr
& {y^{1 - n}}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = \int {\left( {1 - n} \right)Q\left( x \right){e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }}dx} + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Find }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} \cr
& {e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }} = {e^{\int {\left( {1 - 3} \right)\left( {3{x^2}} \right)dx} }} = {e^{ - \int {6{x^2}dx} }} = {e^{ - 2{x^3}}} \cr
& {\text{Substituting }}{e^{\int {\left( {1 - n} \right)P\left( x \right)dx} }},{\text{ }}n,{\text{ and }}Q\left( x \right){\text{ into }}\left( {\bf{1}} \right) \cr
& {y^{1 - 3}}{e^{ - 2{x^3}}} = \int {\left( {1 - 3} \right)\left( {{x^2}} \right)\left( {{e^{ - 2{x^3}}}} \right)dx} + C \cr
& {y^{ - 2}}{e^{ - 2{x^3}}} = \int {2{x^2}{e^{ - 2{x^3}}}dx} + C \cr
& {\text{Integrating}} \cr
& {y^{ - 2}}{e^{ - 2{x^3}}} = \frac{1}{3}\int {\left( { - 6{x^2}} \right){e^{ - 2{x^3}}}dx} + C \cr
& {y^{ - 2}}{e^{ - 2{x^3}}} = \frac{1}{3}{e^{ - 2{x^3}}} + C \cr
& {\text{Simplifying}} \cr
& {y^{ - 2}}{e^{ - 2{x^3}}}{e^{2{x^3}}} = \frac{1}{3}{e^{ - 2{x^3}}}{e^{2{x^3}}} + C{e^{2{x^3}}} \cr
& {y^{ - 2}} = \frac{1}{3} + C{e^{2{x^3}}} \cr
& \frac{1}{{{y^2}}} = \frac{1}{3} + C{e^{2{x^3}}} \cr} $$
