[tex]\dfrac{\mathrm dx}{\mathrm dy}+\dfrac y3=0\iff\mathrm dx=-\dfrac y3\,\mathrm dy[/tex]
[tex]\implies\displaystyle\int\mathrm dx=-\frac13\int\mathrm y\,dy[/tex]
[tex]\implies x=-\dfrac16y^2+C[/tex]
Given that [tex]y(0)=12[/tex], we have
[tex]0=-\dfrac16(12)^2+C\implies C=24[/tex]
So the particular solution to the initial value problem is
[tex]x=-\dfrac16y^2+24\iff y^2=144-6x[/tex]
