Finding the derivative of a relational problem I am self studying some calculus and I have gotten really stuck! I thought I had the right idea but I keep getting the answer totally wrong. I am sure I am missing something important. Here is the problem:
For the equation $6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$, find an equation of the tangent line at the point $(1, 4)$.
Here is my work:
$$6x^{\frac{1}{2}}+12y^{-\frac{1}{2}} = 3xy$$
$$6(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = 3xy$$
$$2(x^{\frac{1}{2}}+2y^{-\frac{1}{2}}) = xy$$
$$2x^{\frac{1}{2}}+4y^{-\frac{1}{2}} = xy$$
$$\ln(2x^{\frac{1}{2}})+\ln(4y^{-\frac{1}{2}}) = \ln(xy)$$
$$\ln2 + \ln x^{\frac{1}{2}}+\ln4 + \ln y^{-\frac{1}{2}} = \ln x + \ln y$$
$$\ln2 + \frac{1}{2}\ln x+\ln4 -\frac{1}{2} \ln y = \ln x + \ln y$$
$$\ln2 + \ln4 + \frac{1}{2}\ln x - \ln x =  \ln y + \frac{1}{2} \ln y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x} = y$$
$$\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x = \ln y$$
$$\frac{1}{3x} = y'\frac{1}{y}$$
$$y' = \frac{y}{3x}$$
$$y' = \frac{e^{\frac{2}{3}\ln2 + \frac{2}{3}\ln4 - \frac{1}{3}\ln x}}{3x}$$
$$y - 4 = \frac{4}{3}(x-1)$$
$$y = \frac{4}{3}x-\frac{4}{3} + 4$$
$$y = \frac{4}{3}x+\frac{8}{3}$$
The Solution I found was: $y = \frac{4}{3}x+\frac{8}{3}$ but this is wrong! Can you tell me where I have gone wrong?

From the fourth line to the fifth, there is a mistake. You assumed that $\ln(a+b)=\ln(a)+\ln(b)$. This is not true.
How to do it:
Start from $6x^{1/2}+12y^{-1/2}=3xy$. It is probably a good idea to do no algebraic manipulation. But let's divide through by $3$, getting $2x^{1/2}+4y^{-1/2}=xy$. 
Don't wait, differentiate, using implicit differentiation.
We get 
$$2\cdot\frac{1}{2}x^{-1/2}-4\cdot\frac{1}{2}y^{-3/2}\frac{dy}{dx}=x\frac{dy}{dx}+y.$$
Now substitute our values of $x$ and $y$ to find $\frac{dy}{dx}$ at the point $(1,4)$.  
