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

$$y'(x) = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
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:

$$y' = {{y - e^{x/y} \over y} \cdot -{y^2 \over e^{x/y} x  - y^2}} = \frac{y(e^{x/y}-y)}{xe^{x/y}-y^2}$$
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