Answer
$$\frac{dy}{dx}=\frac{y(x\ln y-y)}{x(y\ln x-x)}$$
Work Step by Step
$$x^y=y^x$$
First, we take the natural logarithm of both sides: $$\ln(x^y)=\ln(y^x)$$ $$y\ln x=x\ln y$$
Then find $dy/dx$ using implicit differentiation: $$\frac{dy}{dx}\ln x+\frac{y}{x}=\ln y+\frac{x}{y}\frac{dy}{dx}$$ $$\frac{dy}{dx}\Big(\ln x-\frac{x}{y}\Big)=\ln y-\frac{y}{x}$$ $$\frac{dy}{dx}\Big(\frac{y\ln x-x}{y}\Big)=\frac{x\ln y -y}{x}$$ $$\frac{dy}{dx}=\frac{\frac{x\ln y -y}{x}}{\frac{y\ln x-x}{y}}=\frac{y(x\ln y-y)}{x(y\ln x-x)}$$
