Implicit functions and related differential equations I'm seeking guidance in derivation of implicit equation solutions to second degree differential equations.In the example below, differentiating twice just produced a tangle of terms which did not obviously lead to the required result.
Example: If 
$$y^3 +3yx +2x^3 = 0, $$
prove that $$x^2(1+x^3)y'' - (3/2)xy' +y =0$$

Borrow ideas from the numerical variant of Cardano's method for cubic equations.
Set $y=2\sqrt{x}\sinh(u)$, then
\begin{align}
0&=y^3+3xy+2x^3\\
&=2x\sqrt x(4\sinh(u)^3+3\sinh(u))+2x^3\\
&=2x\sqrt x\sinh(3u)+2x^3
\end{align}
$$\\~\\
\implies u=-\frac13\text{asinh}(x^{3/2}),
~~
u'=-\frac12\frac{\sqrt{x}}{\sqrt{1+x^3}}=-\frac12(x^{-1}+x^2)^{-1/2},
~~
u''=\frac14\frac{-x^{-2}+2x}{(x^{-1}+x^2)^{3/2}}
$$
and on the other side
$$
\left(\frac{y}{2\sqrt x}\right)'=\cosh(u)u',~~\left(\frac{y}{2\sqrt x}\right)''=\sinh(u)u'^2+\cosh(u)u''
\\~\\
\implies
u'\left(\frac{y}{2\sqrt x}\right)''=u''\left(\frac{y}{2\sqrt x}\right)'+u'^3\left(\frac{y}{2\sqrt x}\right).
$$
This is a linear ODE in $y$, as $u$ and its derivatives are solved as functions of $x$.
Now insert $u', u''$, apply the quotient rule, and simplify. 

Observing the coefficient structure, one could first collect the terms as
$$
\left(\frac1{u'}\left(\frac{y}{\sqrt x}\right)'\right)'=u'\left(\frac{y}{\sqrt x}\right).
$$

As $(x^{-1/2}y)'=x^{-1/2}y'-1/2x^{-3/2}y=x^{-3/2}(xy'-\frac12y)$, the last formula results in
$$
\left(2\sqrt{x^{-1}+x^{-4}}\left(xy'-\frac12y\right)\right)'
=\frac1{2x\sqrt{1+x^{-3}}}\left(\frac{y}{\sqrt x}\right)
\\~\\
-\frac{x^{-2}+4x^{-5}}{\sqrt{x^{-1}+x^{-4}}}\left(xy'-\frac12y\right)
+2\sqrt{x^{-1}+x^{-4}}\left(xy''+\frac12y'\right)
-\frac1{2x\sqrt{1+x^{-3}}}\left(\frac{y}{\sqrt x}\right)=0
\\~\\
-(x^3+4)\left(xy'-\frac12y\right)
+2x(1+x^{3})\left(xy''+\frac12y'\right)
-\frac{x^3}{2}y=0
\\~\\
2x^2(1+x^3)y''+[x(1+x^3)-x(x^3+4)]y'+2y=0
\\~\\
x^2(1+x^3)y''-\frac32xy'+y=0
$$
which is indeed the claimed differential equation.
