If x = e^(x/y), then prove that (dy)/(dx) = (x-y)/(x log x) (Understanding)

If $x = e^{\frac{x}{y}}$, then prove that $\frac{dy}{dx} = \frac{x-y}{x \log x}$ (Understanding)

Answer

Official answer key

Given $x = e^{x/y}$. Taking logs: $y\log x = x$. Differentiating gives $\frac{dy}{dx} = \frac{x-y}{x\log x}$.

Given $x = e^{x/y}$; take natural logarithm of both sides: $\log x = \frac{x}{y}\log e = \frac{x}{y}$
Multiply by $y$: $y\log x = x$
Differentiate both sides w.r.t. $x$ using the product rule: $y\cdot\frac{1}{x} + \log x\,\frac{dy}{dx} = 1$
Rearrange: $\log x\,\frac{dy}{dx} = 1 - \frac{y}{x} = \frac{x-y}{x}$
Divide by $\log x$: $\frac{dy}{dx} = \frac{x-y}{x\log x}$