Answer
$$y = \frac{{12}}{5}{x^2} + C{x^{ - 3}}$$
Work Step by Step
$$\eqalign{
& 3\left( {y - 4{x^2}} \right)dx + xdy = 0 \cr
& {\text{Divide by }}xdx \cr
& \frac{{3\left( {y - 4{x^2}} \right)dx}}{{xdx}} + \frac{{xdy}}{{xdx}} = 0 \cr
& \frac{{3\left( {y - 4{x^2}} \right)}}{x} + \frac{{dy}}{{dx}} = 0 \cr
& \frac{3}{x}y - 12x + \frac{{dy}}{{dx}} = 0 \cr
& \frac{{dy}}{{dx}} + \frac{3}{x}y = 12x \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = \frac{3}{x}{\text{, }}Q\left( x \right) = 12x \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{\int {\frac{3}{x}dx} }} = {e^{3\ln x}} = {x^3} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {x^3}\left( {\frac{{dy}}{{dx}} + \frac{3}{x}y} \right) = {x^3}\left( {12x} \right) \cr
& {x^3}\frac{{dy}}{{dx}} + 3{x^2}y = 12{x^4} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{x^3}y} \right] = 12{x^4} \cr
& d\left[ {{x^3}y} \right] = 12{x^4}dx \cr
& {\text{Integrate both sides}} \cr
& {x^3}y = \int {12{x^4}} dx \cr
& {x^3}y = \frac{{12}}{5}{x^5} + C \cr
& {\text{Solve for }}y \cr
& y = \frac{{12}}{5}{x^2} + C{x^{ - 3}} \cr} $$
