<s>[INST] 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
$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
 [/INST] 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...
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