Answer
$$\frac{dy}{dx}=\frac{y(1-xe^{x+y})}{x(ye^{x+y}-1)}$$
Work Step by Step
$$\ln xy=e^{x+y}$$
We have $$(\ln x)'=\frac{1}{x}$$ $$(e^{x})'=e^x$$
Carrying out implicit differentiation on both sides, we have $$\frac{1}{xy}\frac{d}{dx}(xy)=e^{x+y}\frac{d}{dx}(x+y)$$
$$\frac{y+x\frac{dy}{dx}}{xy}=e^{x+y}(1+\frac{dy}{dx})$$
$$\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=e^{x+y}+e^{x+y}\frac{dy}{dx}$$
$$e^{x+y}\frac{dy}{dx}-\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}-e^{x+y}$$
$$\Big(e^{x+y}-\frac{1}{y}\Big)\frac{dy}{dx}=\frac{1}{x}-e^{x+y}$$
$$\frac{dy}{dx}=\frac{\frac{1}{x}-e^{x+y}}{e^{x+y}-\frac{1}{y}}=\frac{\frac{1-xe^{x+y}}{x}}{\frac{ye^{x+y}-1}{y}}$$
$$\frac{dy}{dx}=\frac{y(1-xe^{x+y})}{x(ye^{x+y}-1)}$$
