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.

You are almost there.
Note that you found:
$$y'=\frac{-3x}{4y}$$
$$\therefore xyy'=\frac{-3x^2}{4}\tag{1}$$
and from the initial identity we have:$$3x^2=12-4y^2=4(3-y^2)\tag{2}$$
Now just substitute (2) into (1) and use that result to simplify your answer.
