I wish to solve $$y'-\frac{1}{3x}y=\frac{x}{3y}$$ At first, rearrange and simplify to $$y'=\frac{x^2+y^2}{3xy}$$ which clearly indicates that it is a homogeneous differential equation.
Apply the substitution $y=ux\rightarrow y'=u'x+u$ which gives $$u'x=\frac{1-2u^2}{3u}$$ and by integrating we get that $$-\frac{3}{4}\ln\left| 2u^2-1\right| = \ln \left| x\right|+c$$ $$e^{-3/4\ln\left| 2u^2-1\right|} = e^{\ln \left| x\right|+c}$$ $$\left| 2u^2-1\right|^{-3/4}=C \left |x \right|$$ $$\left| 2u^2-1\right|=(C \left |x \right|)^{-4/3}$$ $$\left| 2u^2-1\right|=C' \left |x ^{-4/3}\right|$$ $$\left| 2u^2-1\right|=C' x ^{-4/3}$$ Now, if I choose $ 2u^2-1=C' x ^{-4/3}$ and solve for $u$ and then for $y$, I am getting the expected $y(x) = \pm\sqrt{x^2/2+C' x^{2/3}}$. But, if I choose $ 1-2u^2=C' x ^{-4/3}$, I am getting $y(x) = \pm\sqrt{x^2/2-C' x^{2/3}}$ which are not listed in the answers at least at Alpha.
How to choose between the two? Note that $C'$ is clearly positive.
You didn't solve the second case correctly. If
$$ 1 - 2u^2 = C'x^{-4/3} $$
then
$$ u^2 = \frac12 - \bar{C}e^{-4/3} $$
Thus $$ y = \pm \sqrt{\frac{x^2}{2} - \bar{C}x^{2/3}} $$
$\bar{C}$ is indeed positive in this case. However, the other case flips the sign on $\bar{C}$, so we can combine the two cases by replacing $\pm \bar{C} \mapsto c$, where $c$ can be positive or negative, to get the full solution
$$ y = \pm\sqrt{\frac{x^2}{2} + c x^{2/3}} $$
which is the solution given by WolframAlpha