Second order ODE, first derivative missing I have the following second order equation, where the first derivative is missing, and I am asked to find its general solution:
$$6x^{2}yy''=3x(3y^{2}+2)+2(3y^{2}+2)^3$$
I don't know how to solve it. I have tried with a $u(x)=3y^{2}+2$ substitution but it doesn't seem useful...
Is there any method for this kind of equation whithout $y'$?

Let 
$$ u = 3 y^2 +2, u'= 6 y y'  $$
where primes are with respect to x. We need to solve
$$ y y''= u/(2 x)  + u^3 / (3 x^2) \tag{0} $$

 Spoilt earlier erroneous part, substituted by new derivation. 
 $$ \frac{u'}{6} = \frac{u}{2 x} +  \frac{u^3}{3 x^2} ;$$
 $$ u' = \frac{u}{ x}  ( 3+ 2 \frac{u^2}{x}) ;$$
 Unless $3$ is absent, the variables are not separable. 
 Employing third degree substitution
 $$ u = \frac{ x^3}{\sqrt{  c -4 x^5/5 }} = 3 y^2 +2 ; $$
 is the integrand with arbitrary constant $c ;$

$$ u = 3 y^2 +2  \tag{1} $$
$$ u' = 6 y y' \tag{2} $$
from $ (1), (2) $
$$ \frac{y}{y'}= \frac{ 2(u-2)}{u'} \tag{3}$$
$$ y \, y'' = \frac{u'}{6} \tag{4}$$ 
from $(3),(4) $
$$ y =\sqrt{(u-2)/3}  \tag {5}$$
$$ y'= \frac {u'/2}{\sqrt{3(u-2)}} \tag {6}$$
Differentiate squared equation of (6) and simplify
$$ 24 y' y'' = \frac {2 (u-2) u'u''-u'^3}{(u-2)^2} \tag {7} $$
Eliminate $y'$ from (6) and (7)
$$   y''=  \frac{\sqrt{(u-2)}}{4 {\sqrt 3}u'}  \tag{8}$$
Multiply (5) and (8) and simplify
$$ 12 y y''= 2 u''- \frac{u^{'2}}{u-2} \tag{9}$$
that leads to a second order non-linear ODE in $u$ and $x$.
$$ 2 u''= 12\, (\frac{u}{2 x} + \frac {u^3} {3 x^2} ) + \frac{u'^{2}}{(u-2)} \tag {10}. $$
