Solve this

Here,

$e^{x}+e^{y}=e^{x+y} \ldots \ldots$ (i)

Differentiating both the sides using chain rule,

$\frac{d}{d x} e^{x}+\frac{d}{d x} e^{y}=\frac{d}{d x}\left(e^{x+y}\right)$

$e^{x}+e^{y} \frac{d y}{d x}=e^{x+y} \frac{d}{d x}(x+y)$

$e^{x}+e^{y} \frac{d y}{d x}=e^{x+y}\left[1+\frac{d y}{d x}\right]$

$e^{y} \frac{d y}{d x}-e^{x+y} \frac{d y}{d x}=e^{x+y}-e^{x}$

$\frac{d y}{d x}\left(e^{y}-e^{x+y}\right)=e^{x+y}-e^{x}$

$\frac{d y}{d x}=\frac{e^{x+y}-e^{x}}{e^{y}-e^{x+y}}$

$\frac{d y}{d x}=\frac{e^{x}+e^{y}-e^{x}}{e^{y}-\left(e^{x}+e^{y}\right)}$

$\frac{d y}{d x}=\frac{e^{x}+e^{y}-e^{x}}{e^{y}-e^{x}-e^{y}}$

$\frac{d y}{d x}=\frac{e^{y}}{-e^{x}}$

$\frac{d y}{d x}=-e^{y-x}$

$\frac{d y}{d x}+e^{y-x}=0$

Hence Proved.