How to implicitly differentiate $xy=2e^{x+y-3}$ I am having problems with this question; I can get close to an end result but never all the way. 
What I have done so far is:
$$\frac{xy}2 =e^{x+y-3}$$
and then log both sides
$$\ln{\frac{xy}2}=\ln{e^{x+y-3}}$$
which then gives me
$$\ln{\frac{xy}2}=x+y-3$$
Here is where I think I am making an error with the differentiation
$$\frac2{xy}\frac{y+x\frac d{dx}y}2 = x+y-3$$
$$\frac{y+x\frac d{dx}y}{xy}=x+y-3$$
I'm not sure what to do from here. Any help would be very much appreciated.
 A: Actually, Most of your work was fine; you just stopped too early (and forgot to also implicitly the left side of the equation!). Life can be easier when we exploit the properties of the logarithm, but the problem here doesn't require that they be used.
I'll use $y'$ to denote $\frac{dy}{dx}$.
We have $$\ln{\frac{xy}2}=\underbrace{x+y-3}_{\frac d{dx}\left(x+y-3\right) = 1+\frac{dy}{dx}}$$
You correctly found that:
(1) Using implicit differentiation, we get $\frac{2}{xy}\cdot \frac 12(x y'+ y) = 1+y'$
$$ \iff \frac{xy' + y}{xy} - (1+ y') = 0$$
From there, it's all algebra: 
$$\iff \frac{y'}y + \frac 1x -1 -y'=0 \iff y'\left(\frac 1y -1\right)= 1-\frac 1x$$
$$\iff y'= \frac{1-\frac 1x}{1-\frac 1y} \iff  y' = \frac{dy}{dx} = \frac{y(x-1)}{x(1-y)}$$
A: Following my suggested tips concerning logarithms, we find that if $x,y>0$,
$$\ln x+\ln y-\ln2=x+y-3$$
Differentiate both sides to get
$$\frac1x+\frac1y\frac{dy}{dx}=1+\frac{dy}{dx}$$
And now some algebra:
$$\left(\frac1y-1\right)\frac{dy}{dx}=1-\frac1x$$
$$\frac{dy}{dx}=\frac{1-\frac1x}{\frac1y-1}=\frac{y(x-1)}{x(1-y)}$$

The result is the same if $x,y<0$, but if $x,y$ have different signs, you can't take the log of both sides as you did.  I may note directly differentiating the original problem isn't too hard:
$$\frac d{dx}\frac{xy}2=\frac12\left(y+x\frac{dy}{dx}\right)$$
$$\frac d{dx}e^{x+y-3}=(x+y-3)'e^{x+y-3}=\left(1+\frac{dy}{dx}\right)\frac{xy}2$$
And the solution is the same, justifying the case of $x,y$ with different signs.
