Given

$3t{x}^{\prime}-2x=\frac{{t}^{3}}{{x}^{2}}$

$3t\frac{dx}{dt}-2x=\frac{{t}^{3}}{{x}^{2}}$

$\frac{dx}{dt}-\frac{2x}{3t}=\frac{{t}^{3}}{{x}^{23}t}$

$\frac{dx}{dt}-\frac{2x}{3t}=\frac{{t}^{2}}{3{x}^{2}}$

${x}^{2}\frac{dx}{dt}-\frac{2{x}^{3}}{3t}={t}^{2}v$

$3{x}^{2}\frac{dx}{dt}-\frac{2{x}^{3}}{t}=3{t}^{2}$

Let ${x}^{3}=u$

$3{x}^{2}\frac{dx}{dt}=\frac{du}{dt}$

$\frac{du}{dt}-\frac{2}{t}u=e{t}^{2}$

$\frac{du}{dt}+Pu=Q$ where

$P=\frac{-2}{t};\text{}\text{}Q=3{t}^{2}$

$If={e}^{\int Pdt}={e}^{\int \frac{-2}{t}dt}={e}^{-2\mathrm{ln}t}=\frac{1}{{t}^{2}}$

Solution:

$u\left(If\right)=\int Q\left(If\right)dt+c$

${x}^{3}\frac{1}{{t}^{2}}=\int 3{t}^{2}\frac{1}{{t}^{2}}dt+c=3t+c$

$\frac{{x}^{3}}{{t}^{2}}=3t+c$