Implicit Differentiation $y''$ I'm trying to find $y''$ by implicit differentiation of this problem: $4x^2 + y^2 = 3$
So far, I was able to get $y'$ which is $\frac{-4x}{y}$
How do I go about getting $y''$? I am kind of lost on that part. 

You have $$y'=-\frac{4x}y\;.$$ Differentiate both sides with respect to $x$:
$$y''=-\frac{4y-4xy'}{y^2}=\frac{4xy'-4y}{y^2}\;.$$
Finally, substitute the known value of $y'$:
$$y''=\frac4{y^2}\left(x\left(-\frac{4x}y\right)-y\right)=-\frac4{y^2}\cdot\frac{4x^2+y^2}y=-\frac{4(4x^2+y^2)}{y^3}\;.$$
But from the original equation we know that $4x^2+y^2=3$, so in the end we have
$$y''=-\frac{12}{y^3}\;.$$
