[tex]~~~~~~\ln(xy)= e^{x+y}\\\\\\\implies \dfrac{d}{dx} \ln(xy) = \dfrac{d}{dx}\left(e^{x+y} \right)\\\\\\\implies \left(\dfrac 1{xy}\right) \dfrac{d}{dx}(xy) =\left( e^{x+y} \right)\dfrac{d}{dx} (x+y)~~~~~~~~~~~~~~~~~;[\text{Chain rule}]\\\\\\\implies \left( \dfrac 1{xy} \right) \left(y + x\dfrac{dy}{dx} \right) =\left(e^{x+y}\right) \left( 1 + \dfrac{dy}{dx} \right) \\\\\\[/tex]
[tex]\implies \dfrac 1x + \dfrac 1y\cdot \dfrac{dy}{dx} = e^{x+y}+ \left(e^{x+y} \right) \dfrac{dy}{dx}\\\\\\\implies y + x \dfrac{dy}{dx} = xy e^{x+y}+ \left(xy e^{x+y} \right)\dfrac{dy}{dx}~~~~~~~~~;[\text{Multiply both sides by}~ xy]\\\\\\[/tex]
[tex]\implies x \dfrac{dy}{dx} -\left(xye^{x+y} \right) \dfrac{dy}{dx}= xye^{x+y} -y\\\\\\\implies \left(x - x y e^{x+y} \right) \dfrac{dy}{dx} = xye^{x+y} - y\\\\\\\implies \dfrac{dy}{dx} = \dfrac{xye^{x+y} - y}{x - x y e^{x+y}}\\\\\\\implies \dfrac{dy}{dx} = \dfrac{y \left(xe^{x+y} -1 \right)}{x \left(1- ye^{x+y} \right)}[/tex]
