Answer:[tex]y'=\frac{y^2-xy\ln(y)}{x^2-xy\ln(x)}[/tex]Step-by-step explanation:Take natural log of both sides first.[tex]x^y=y^x[/tex][tex]\ln(x^y)=\ln(y^x)[/tex]Taking the natural log of both sides allows you to bring down the powers.[tex]y\ln(x)=x\ln(y)[/tex]I'm going to differentiate both sides using the power rule.[tex](y)'(\ln(x))+(\ln(x))'y=(x)'(\ln(y))+(\ln(y))'x[/tex]Now recall (ln(x))'=(x)'/x=1/x while (ln(y))'=(y)'/y=y'/y.[tex]y'(\ln(x))+\frac{1}{x}y=1(\ln(y))+\frac{y'}{y}x[/tex]Simplifying a bit:[tex]y' \ln(x)+\frac{y}{x}=\ln(y)+\frac{y'}{y}x[/tex]Now going to gather my terms with y' on one side while gathering other terms without y' on the opposing side.Subtracting y'ln(x) and ln(y) on both sides gives:[tex]\frac{y}{x}-\ln(y)=-y'\ln(x)+\frac{y'}{y}x[/tex]Now I'm going to factor out the y' on the right hand side:[tex]\frac{y}{x}-\ln(y)=(-\ln(x)+\frac{x}{y})y'[/tex]Now we get to get y' by itself by dividing both sides by (-ln(x)+x/y):[tex]\frac{\frac{y}{x}-\ln(y)}{-\ln(x)+\frac{x}{y}}=y'[/tex]Now this looks nasty to write mini-fractions inside a bigger fraction.So we are going to multiply top and bottom by xy giving us:[tex]\frac{y^2-yx\ln(y)}{-xy\ln(x)+x^2}=y'[/tex][tex]y'=\frac{y^2-xy\ln(y)}{x^2-xy\ln(x)}[/tex]