x^y = y^x

The first step is to take the natural log of both sides:

$ln(x^y) = ln(y^x) \therefore yln(x) = xln(y)$

Taking the derivative of both sides with respect to x, using the product and chain rules, gets:

$y \frac{d}{dx}ln(x) + ln(x){dy}{dx}= x \frac {d}{dx}ln(y) + ln(y){d}{dx}x$

$\frac{y}{x} + ln(x) \frac{dy}{dx}= x \frac {d}{dy}ln(y) \frac{dy}{dx} + ln(y)$

$\frac{y}{x} + ln(x) \frac{dy}{dx} = \frac{x}{y} \frac {dy}{dx} + ln(y)$

$\frac{dy}{dx} = \frac{ln(y) - \frac{y}{x}}{ln(x) - \frac{x}{y}}$