The equation is linear, as you can write
[tex]3xy'+y=12x\implies y'+\dfrac1{3x}y=4[/tex]
An integrating factor would be
[tex]\mu(x)=\exp\left(\displaystyle\int\frac{\mathrm dx}{3x}\right)=x^{1/3}[/tex]
Multiplying both sides of the ODE by this yields
[tex]x^{1/3}y'+\dfrac1{3x^{2/3}}y=4x^{1/3}[/tex]
where the LHS is a derivative:
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[x^{1/3}y\right]=4x^{1/3}[/tex]
Integrating both sides, we get
[tex]x^{1/3}y=4\displaystyle\int x^{1/3}\,\mathrm dx[/tex]
[tex]x^{1/3}y=3x^{4/3}+C[/tex]
[tex]y=3x^{3/3}+Cx^{-1/3}=3x+\dfrac C{x^{1/3}}[/tex]
