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