If `e^x+e^y=e^(x+y)` , prove that `(dy)/(dx)=-(e^x(e^y-1))/(e^y(e^x-1))` or, `(dy)/(dx)+e^(y-x)=0`

Explanation

`e^{x}+e^{y}=e^{x+y}`


`Rightarrow e^{x}+e^{y} frac{d y}{d x}=e^{x+y}(1+frac{d y}{d x})`


`Rightarrow e^{x}+e^{y} frac{d y}{d x}=e^{x+y}+e^{x+y} frac{d y}{d x}`


`Rightarrow e^{y} frac{d y}{d x}-e^{x+y} frac{d y}{d x}=e^{x+y}-e^{x}`


`Rightarrow frac{d y}{d x}(e^{y}-e^{x+y})=e^{x+y}-e^{x}`


`Rightarrow frac{d y}{d x}=frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}`


`=frac{e^{x}(e^{y}-1)}{e^{y}(1-e^{x})}`


`=-frac{e^{x}(e^{y}-1)}{e^{y}(e^{x}-1)}`