How do you differentiate ${{e}^{\dfrac{x}{y}}}=x-y$?

Hint: From the question we are asked to differentiate the equation ${{e}^{\dfrac{x}{y}}}=x-y$. So, to solve the question we will use the logarithms and differentiation and its formulae which we will discuss below. First we will apply log on both sides of the equation and proceed the further calculation using the differentiation formulae like $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, $\ln {{x}^{n}}=n\ln x$ and \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] etc.

Complete step by step solution:
We know that $\ln e=1$. So, the equation will be simplified as follows.
First we will apply log on both sides of the equation and proceed the further calculation using the differentiation formulae like $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$, $\ln {{x}^{n}}=n\ln x$and \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] etc.

Firstly, we will take logarithm on both sides of the equation. So, the equation will be reduced as follows.
$\Rightarrow {{e}^{\dfrac{x}{y}}}=x-y$
$\Rightarrow \ln {{e}^{\dfrac{x}{y}}}=\ln \left( x-y \right)$
Here we will use the formula in logarithms which is $\ln {{x}^{n}}=n\ln x$. So, the equation will be reduced as follows.
$\Rightarrow \dfrac{x}{y}\ln e=\ln \left( x-y \right)$
$\Rightarrow \dfrac{x}{y}=\ln \left( x-y \right)$
$\Rightarrow x=y\ln \left( x-y \right)$
Now, we will differentiate the above equation. Then we get,
$\Rightarrow \dfrac{d}{dx}\left( x \right)=\dfrac{d}{dx}\left( y\ln \left( x-y \right) \right)$
Here we use the formulae $\dfrac{d}{dx}\left( x \right)=1$ and $\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ along with the UV rule in differentiation which is \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]. So, the equation after using these formulae will be simplified as follows.
$\Rightarrow 1=\dfrac{y}{x-y}\left( 1-\dfrac{dy}{dx} \right)+\ln \left( x-y \right)\dfrac{d}{dx}\left( y \right)$
Now, we will simplify the equation by multiplying the \[\dfrac{y}{x-y}\] with \[\left( 1-\dfrac{dy}{dx} \right)\]. So, the equation after doing the simplification will be as follows.
$\Rightarrow 1=\dfrac{y}{x-y}-\dfrac{y}{x-y}.\dfrac{dy}{dx}+\ln \left( x-y \right)\dfrac{d}{dx}\left( y \right)$
$\Rightarrow 1-\dfrac{y}{x-y}=\left( \ln \left( x-y \right)-\dfrac{y}{x-y} \right)\dfrac{d}{dx}\left( y \right)$
$\Rightarrow x-2y=\left[ \left( x-y \right)\ln \left( x-y \right)-y \right]\dfrac{dy}{dx}$
Here we will make the required \[\dfrac{dy}{dx}\] as subject and send the remaining terms to one side of the equation. So, the equation will be reduced as follows.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{x-2y}{\left( x-y \right)\ln \left( x-y \right)-y}$
Since, $\Rightarrow {{e}^{\dfrac{x}{y}}}=x-y$ and $\Rightarrow \dfrac{x}{y}=\ln \left( x-y \right)$, we can further simplify the equation as follows.
$\Rightarrow \dfrac{dy}{dx}=\dfrac{xy-2{{y}^{2}}}{x{{e}^{\dfrac{x}{y}}}-{{y}^{2}}}$.
Therefore, the solution will be $\Rightarrow \dfrac{dy}{dx}=\dfrac{xy-2{{y}^{2}}}{x{{e}^{\dfrac{x}{y}}}-{{y}^{2}}}$.

Note: Students must be very careful in calculations. Students must have good knowledge in concepts of differentiation and logarithms. Students must have grip in the differentiation formulae like,
$\dfrac{d}{dx}\left( \ln x \right)=\dfrac{1}{x}$ and also most important product rule or called UV rule which is \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]. Students should not do mistake in using the formula like if we use \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}-v\dfrac{du}{dx}\] instead of \[\Rightarrow \dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] our whole solution will be wrong.