Finding the second derivative implicitely 
$$\text{Having the equation  }\sqrt{y}+3xy=4 \text{ find } \frac{d^2y}{dx^2} := y''$$

Ok so when I work it out implicitly I get, 
$$y'=\frac{-6y^{3/2}}{1+6x\sqrt{y}}$$
Continuing on I get,
$$y''=\frac{90y^2(1+6x\sqrt{y})-108xy^{3/2}}{(1+6x\sqrt{y})^3}$$
Which is incorrect I believe. Can someone please tell me if my $y'$ is correct? And where should I go from there?
 A: If you have the equation (considering that $y = y(x)$ and that $y' := dy/dx$) 

$$y^{1/2} + 3xy = 4 \implies \frac{1}{2y^{1/2}}y' + 3y + 3xy' = 0 \implies y' = \frac{-6y^{3/2}}{1 + 6xy^{1/2}} $$

Where I differentiate both sides of the equation and isolated $y'$. So your answer for $y'$ is correct.

Now lets calculate the second derivative $y''$
$$y'' \stackrel{\text{Leibnitz rule}}{=} \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - (-6y^{3/2})\left(6y^{1/2} + \frac{3xy'}{y^{1/2}}\right)\right) \implies $$
$$y'' = \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - \left(\frac{-6y^{3/2}}{y^{1/2}}\right)\left(6y + 3xy'\right)\right) \implies $$
$$y'' = \frac{1}{(1 + 6xy^{1/2})^2}\left((1 + 6xy^{1/2})(-9y^{1/2}y') - \left(-6y\right)\left(\frac{6yy'}{y'} + 3xy'\right)\right) \implies $$
$$y'' = \frac{y'}{(1 + 6xy^{1/2})^2}\left(-9y^{1/2} -54xy + 36\frac{y^2}{y'} +18xy \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy + 36y^2\frac{(1+6xy^{1/2})}{-6y^{3/2}}\right) \implies $$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy -6y^{1/2}(1+6xy^{1/2}) \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-9y^{1/2} - 36xy -6y^{1/2} -36xy \right) \implies$$
$$y'' = \frac{-6y^{3/2}}{(1 + 6xy^{1/2})^3}\left(-15y^{1/2} - 72xy\right) \implies$$
Which leads us to the answer

$$y'' = \frac{90y^2+432xy^{5/2}}{(1 + 6xy^{1/2})^3}$$

A: With some serious careful simplification the answer I finally worked out to be 
$$y''=\frac{90y^2(1+6x\sqrt{y})-108xy^{5/2}}{(1+6x\sqrt{y})^3}$$
Note the ever so slight difference in my posted question. 
Very annoying problem to say the least.
A: $y=(4-3xy)^2=>y=16-9x^2y^2-24xy$ now you need to use the product rule when differentiating the $-9x^2y^2$ term and $-24xy$
$\frac{dy}{dx}=\frac{-18xy^2-24y}{-1+18x^2y-24x}$
