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.