Given $e^{x+y} = xy$
Taking log on both sides, we get: $ (x+y) \log_e e = \log_e x + \log_e y$
$\Rightarrow x+y = \log_e x + \log_e y$
Differentiating, we get: $dx + dy = \large\frac{1}{x}$$dx + \large\frac{1}{y}$$dy$
Rearranging, we get: $\large\frac{dy}{dx}$$(1 - \large\frac{1}{y}$$) = \large\frac{1}{x}$$-1$
$\Rightarrow \large\frac{dy}{dx} $$ = \large\frac{y(1-x)}{x(y-1)}$