Find \( \frac{d y}{d x} \) by implicit differentiation. \[ e^{x / y}=5 x-y \] \( \frac{d y}{d x}=\square \)

Differentiate both sides of \[ e^{x/y}=5x-y \] with respect to \(x\). 1. On the left-hand side, let \[ u=\frac{x}{y}. \] Then \[ \frac{d}{dx}\left(e^{u}\right)=e^{u}\frac{du}{dx}. \] Since \[ u=\frac{x}{y}=x\cdot y^{-1}, \] its derivative is \[ \frac{du}{dx}=\frac{d}{dx}\left(x\cdot y^{-1}\right) = \frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}. \] Thus, the derivative of the left-hand side is \[ e^{x/y}\left(\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right). \] 2. Differentiate the right-hand side: \[ \frac{d}{dx}(5x-y)=5-\frac{dy}{dx}. \] 3. Equate the two derivatives: \[ e^{x/y}\left(\frac{1}{y}-\frac{x}{y^2}\frac{dy}{dx}\right)=5-\frac{dy}{dx}. \] 4. Distribute and rearrange to isolate \(\frac{dy}{dx}\): \[ \frac{e^{x/y}}{y}-\frac{x e^{x/y}}{y^2}\frac{dy}{dx}=5-\frac{dy}{dx}. \] Bring all terms containing \(\frac{dy}{dx}\) to one side: \[ -\frac{x e^{x/y}}{y^2}\frac{dy}{dx} + \frac{dy}{dx}=5-\frac{e^{x/y}}{y}. \] Factor out \(\frac{dy}{dx}\): \[ \frac{dy}{dx}\left(1-\frac{x e^{x/y}}{y^2}\right)=5-\frac{e^{x/y}}{y}. \] 5. Now, solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx}=\frac{5-\frac{e^{x/y}}{y}}{1-\frac{x e^{x/y}}{y^2}}. \] This is the derivative in its acceptable form. Alternatively, one can multiply numerator and denominator by \(y^2\) to clear the fractions: \[ \frac{dy}{dx}=\frac{5y^2-e^{x/y}y}{y^2-x e^{x/y}}. \] Thus, the final answer is: \[ \frac{dy}{dx}=\frac{\frac{e^{x/y}}{y}-5}{\frac{x e^{x/y}}{y^2}-1} \quad\text{or}\quad \frac{dy}{dx}=\frac{y\;e^{x/y}-5y^2}{x\;e^{x/y}-y^2}. \]