Implicit differentiation of $3x^2 + 4y^2 = 12$ 
The equation $3x^2 + 4y^2 = 12$ defines $y$ implicitly as two functions of $x$ if $|x| < 2$ and $|x|=2$. Assuming
  the second derivative $y''$ exists, show that it satisfies the equation $4y^3 y'' = -9$.

Progress.  I derived $6x+8yy'=0$ then $y'= \frac{-3x}{4y}$ then derived again $$y'' = \frac{-3(y-xy')}{4y^2}= \frac{-3(y-xy')y}{4y^2y}= \frac{-3(y^2-xyy')}{4y^3} $$ Then $4y^3y''= -3(y^2-xyy')$, so I want $-9 = -3(y^2-xyy')$... and don't know the rest.

It turns out that
if
$ax^2+by^2 = c$,
then
$b^2y^3y''
=-ac
$.
In this case,
$16y^2y''
= -36$
or
$4y^2y''
= -9$.
I use implicit differentiation
twice.
From
$ax^2 + by^2 = c
$,
as you did
$0 
=2ax+2byy'
=ax+byy'
$,
so
$y'
=-\frac{ax}{by}
$.
Doing it again,
$0
= a+b((y')^2+yy'')
$
or
$-\frac{a}{b}
=(y')^2+yy''
=(\frac{ax}{by})^2+yy''
$
or
$-aby^2
=a^2x^2+b^2y^3y''
$
or
$b^2y^3y''
=-aby^2-a^2x^2
=-a(by^2+ax^2)
=-ac
$.
In your case,
$c = ab$,
so
$b^2y^3y''
=-a^2b
$
or
$by^3y''
=-a^2
$.
