Answer
$$y= x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) .$$
Work Step by Step
Rewrite the equation
$$y'-(3x)^{-1}y=\frac{1}{3}x^{-2}$$
This is a linear equation and has the integrating factor as follows
$$\alpha(x)= e^{\int P(x)dx}=e^{\int -(3x)^{-1} dx}=e^{- \frac{1}{3}\ln x}=x^{-\frac{1}{3}}$$
Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=x^{\frac{1}{3}}\left( \int \frac{1}{3}x^{-\frac{7}{3}} dx +C\right)\\ &=x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) \end{align}
Thus, the general solution is
$$y= x^{\frac{1}{3}}\left( -\frac{1}{4} x^{-4/3}+C\right) .$$
