Answer
$-(2+\ln 2)$
Work Step by Step
Since, we have $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ ...(1)
Now, need to plug the derivatives in equation (1).
we have $\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1} \times (-1)=\dfrac{-e^y -y \cos xy}{xe^y+x \cos xy+1} $
Now, plug the point $(0,\ln 2)$
Since, $e^{\ln 2}=2$
Thus, we get $\dfrac{-(e^{\ln 2} +(\ln 2) \cos (0) (\ln 2))}{0+0+1} =-\dfrac{2+\ln 2}{1} =-(2+\ln 2)$
