## Thomas' Calculus 13th Edition

$-(2+\ln 2)$
Consider $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ Plug the derivatives in equation $\dfrac{dy}{dx}=-\dfrac{F_x}{F_y}$ This implies that $\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1} (-1)=-\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1}$ For the point $(0,\ln 2)$ As we know that $e^{\ln 2}=2$ So, $-\dfrac{e^y+y \cos xy}{xe^y+x \cos xy+1}=-\dfrac{e^{\ln 2} +(\ln 2) \cos (0) (\ln 2)}{1} =-(2+\ln 2)$