If $$e^x+e^y=e^{x+y}$$, show that $$\dfrac{dy}{dx}=-e^{y-x}$$
Given
$$e^x+e^y=e^{x+y}$$...........(i)
differentiating w.r to x
$$e^x+e^y.\dfrac{dy}{dx}=e^{x+y}(1+\dfrac{dy}{dx})$$
$$e^x+e^y.\dfrac{dy}{dx}=e^{x+y}+e^{x+y}.\dfrac{dy}{dx}$$
$$\dfrac{dy}{dx}(e^y-e^{x+y})=e^{x+y}-e^x$$
$$\dfrac{dy}{dx}=\dfrac{e^{x+y}-e^x}{e^y-e^{x+y}}$$
Put the value of $$e^{x+y}$$ from eqn. (1)
$$\therefore \dfrac{dy}{dx}=\dfrac{e^x+e^y-e^x}{e^y-e^x-e^y}$$
$$\dfrac{dy}{dx}=\dfrac{e^y}{-e^x}$$
$$\dfrac{dy}{dx}=-e^{y-x}$$
Hence proved