[tex]y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}[/tex]
Divide both sides by [tex]\dfrac13y^{-2}(x)[/tex]:
[tex]3y^2y'-y^3=xe^x\ln x[/tex]
Substitute [tex]v(x)=y(x)^3[/tex], so that [tex]v'(x)=3y(x)^2y'(x)[/tex].
[tex]v'-v=xe^x\ln x[/tex]
Multiply both sides by [tex]e^{-x}[/tex]:
[tex]e^{-x}v'-e^{-x}v=x\ln x[/tex]
The left side can be condensed into the derivative of a product.
[tex](e^{-x}v)'=x\ln x[/tex]
Integrate both sides to get
[tex]e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C[/tex]
Solve for [tex]v(x)[/tex]:
[tex]v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]
Solve for [tex]y(x)[/tex]:
[tex]y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x[/tex]
[tex]\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}[/tex]
