What is the implicit differnation of $e^\frac{x}{y}=(x-y)$ I am trying to solve this derivate but i  am really lost.
Its an implicit function 
$$e^\frac{x}{y}=(x-y)$$
I have a feeling its chain rule but i am really lost on how to apply it. Maybe someone can give me a helping hand
 A: You have $$e^{x/y} = x-y$$ Hence,
\begin{align}
\dfrac{d(e^{x/y})}{dx} & = \dfrac{d(x-y)}{dx}\\
\underbrace{\dfrac{d (e^{x/y})}{d(x/y)} \times \dfrac{d(x/y)}{dx}}_{\text{Chain rule}} & = \dfrac{dx}{dx} - \dfrac{dy}{dx}\\
e^{x/y} \times \dfrac{d(x/y)}{dx} & = 1 - \dfrac{dy}{dx}\\
\end{align}
Now lets evaluate $\dfrac{d(x/y)}{dx}$. We have that
$$\dfrac{d(x/y)}{dx} = \underbrace{\dfrac1y \dfrac{dx}{dx} + x \dfrac{d(1/y)}{dx}}_{\text{Product rule}} = \dfrac1y + x\underbrace{\dfrac{d(1/y)}{dy} \dfrac{dy}{dx}}_{\text{Chain rule}} = \dfrac1y - \dfrac{x}{y^2} \dfrac{dy}{dx}$$
Putting this together, we get that
$$e^{x/y} \times \left(\dfrac1y - \dfrac{x}{y^2} \dfrac{dy}{dx} \right) = 1 - \dfrac{dy}{dx}$$
Rearranging terms and collection terms containing $\dfrac{dy}{dx}$, we get that
$$\left(1 - \dfrac{x}{y^2}e^{x/y} \right) \dfrac{dy}{dx} = 1 - \dfrac{e^{x/y}}y$$
Hence, we get that
$$\dfrac{dy}{dx} = \dfrac{\left(1 - \dfrac{e^{x/y}}y \right)}{\left(1 - \dfrac{x}{y^2}e^{x/y} \right)}$$
A: I'll do the left-hand side.
$$ \frac{d(e^{x/y})}{dx} = e^{x/y} \frac{d(x/y)}{dx} = e^{x/y}(\frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx})$$
In the first step, I used the chain rule, so I took the deriative of the exponential and then the derivative of the exponent $x/y$. To take the derivative of $x/y$ I applied the product rule to $xy^{-1}$, and lastly $dy/dx$ appears when you apply the chain rule to the derivative of $y^{-1}(x)$, remembering that $y$ is a function of $x$.
