Answer
$$-(\ln2+1)$$
Work Step by Step
Re-write as: $$2\times (xy)+e^{x+y}-2=0 \\2y \ dx+2x\ dy+e^{x+y}\ dx+e^{x+y}dy=0$$
When $x=0,y=\ln2$, then we have:
$2\ln2dx+e^{\ln2}dx+e^{\ln2}dy=0 \\2\ln2dx+2dx+2dy=0 \\-(2\ln2+2)dx =2dy\\ \dfrac{dy}{dx}=-(\ln2+1)$$
