Is this the best route to go in this implicit derivation problem? I have this equation:
$$e^{\frac{x}{y}} = x - y$$
I seem to be going down the wrong path. Is this right so far?
$$\frac{dy}{dx} = e^{\frac{x}{y}} \cdot \frac{dy}{dx} (x \cdot y^{-1}) = 1 - y'$$
$$e^{\frac{x}{y}} ( x \cdot -y^{-2} \cdot y' + y^{-1}) = 1 - y'$$
$$ -e^{\frac{x}{y}} x \cdot y^{-2} \cdot y' + e^{\frac{x}{y}} y^{-1}) = 1 - y'$$
WolframAlpha has this as
$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
 A: I don't know why your subtraction sign became an equals sign, but your goal, after implicitly differentiating, is to isolate the $y'$ terms so you can factor and divide:
$$e^{x/y} \left( x \cdot -y^{-2} \cdot y' + y^{-1}\right) = 1 - y'$$
$$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + e^{x/y} \cdot y^{-1} = 1 - y'$$
$$e^{x/y} \cdot x \cdot -y^{-2} \cdot y' + y' = 1 - e^{x/y} \cdot y^{-1}$$
Then factor:
$$y'\left(e^{x/y} \cdot x \cdot -y^{-2} + 1\right) = 1 - e^{x/y} \cdot y^{-1}$$
And divide. After some simplification, you should reach:
$$y' = {{y - e^{x/y} \over y} \cdot -{y^2 \over e^{x/y} x  - y^2}} = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
A: Your result $$-e^{\frac{x}{y}}xy^{-2}\color{blue}{y'}+e^{\frac{x}{y}}y^{-1}=1-\color{blue}{y'}$$ is correct. 
You can continue and simplify this by collecting all the terms with $y'$ as a factor:
$$\begin{align}\color{blue}{y'}\left (1-e^{\frac{x}{y}}xy^{-2}\right )=1-e^{\frac{x}{y}}y^{-1}\end{align}\\\color{blue}{y'}=\frac{1-e^{\frac{x}{y}}y^{-1}}{1-e^{\frac{x}{y}}xy^{-2}}\frac{y^2}{y^2}=\frac{y(y-e^{\frac{x}{y}})}{y^2-e^{\frac{x}{y}}x}$$
This is the same answer that WolframAlpha gave you, only that they have multiplied numerator and denominator by $-1$.
A: Why is nobody suggesting logarithmic differentiation ?
First take log base $e$ of both sides to get 
$$\frac{x}{y(x)}=\ln (x-y(x))\Rightarrow x=y(x)\cdot \ln (x-y(x)).$$
Now differentiate implicitly:
$$\begin{align}
1&=y\cdot \frac{1}{x-y}\cdot \left(1-\frac{dy}{dx}\right)+\ln(x-y)\cdot \frac{dy}{dx}
\\ \Rightarrow 1&=\frac{y}{x-y}-\frac{y}{x-y}\cdot \frac{dy}{dx}+\ln(x-y)\cdot \frac{dy}{dx}
\\ \Rightarrow \frac{dy}{dx}\left(\ln(x-y)-\frac{y}{x-y}\right)&=\frac{x-2y}{x-y}
\\ \Rightarrow \frac{dy}{dx}&=\frac{\frac{x-2y}{x-y}}{\ln(x-y)-\frac{y}{x-y}}
\\ \Rightarrow \frac{dy}{dx}&=\frac{x-2y}{(x-y)\ln(x-y)-y}.
\end{align}$$
OK, maybe now I know why...
A: We have 
$$
f(x,y) = e^{\frac{x}{y}}-x+y=0
$$
so we have
$$
df = f_x dx + f_y dy = 0\Rightarrow \frac{dy}{dx}=-\frac{f_x}{f_y} = \frac{y (x-2 y)}{x^2-x y-y^2}
$$
A: Please check your line 2, it should be something like this:
$$e^{\frac{x}{y}}\times(y^{-1}-xy^{-2}y')=1-y'$$
Proceed using this to get the WolframAlpha  output
A: $e^{\frac{x}{y}} = x - y$
differentiate wrt $y$ and use the chain rule and Quotient rule on $e^{\frac xy}$ in that order ;
$ e^{\frac xy}\cdot \frac{y-\frac{dy}{dx}x}{y^2}=1-\frac{dy}{dx} $
$e^{\frac xy} y -\frac{dy}{dx}x e^{\frac xy} = y^2-y^2\frac{dy}{dx}$
$\frac{dy}{dx}\big(y^2-x\cdot e^{\frac xy}\big) = y^2-e^{\frac xy}y$
$\frac{dy}{dx} = \frac{y(y-e^{\frac xy})}{y^2-x\cdot e^{\frac xy}}$
