Find particular solution of the separable equation
$\displaystyle 6(x^4-3x)dx+\frac{1}{y^3}dy=0$
$\displaystyle (6x^4-18x)dx=(-y^{-3})dy$
$\displaystyle \int(6x^4-18x)dx=-\int(y^{-3})dy$
$\displaystyle \frac{6}{5}x^5-9x^2+c=-\frac{-y^-2}{-2}$
$\displaystyle \frac{6}{5}x^5-9x^2+c=\frac{1}{2y^2}$
Looks right except for that you need a 2 on the final step
$\displaystyle (6x^4-18x)dx=(-y^-3)dy$
$\displaystyle \int(6x^4-18x)dx=\int(-y^-3)dy$
$\displaystyle \frac{6}{5}x^5-9x^2+c=\frac{1}{2y^2}$
$\displaystyle y(0)=1$
$\displaystyle \frac{6}{5}(0)^5-9(0)^2+c=\frac{1}{2(1)^2}$
$\displaystyle c=\frac{1}{2}$
Now I have:
$\displaystyle \frac{6}{5}x^5-9x^2+\frac{1}{2}=\frac{1}{2y^2}$