## Thomas' Calculus 13th Edition

$2$
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 $\dfrac{(y-3)}{(x+2y)} (-1)=\dfrac{3-y}{(x+2y)}$ For point $(-1,1)$, we have $\dfrac{3-y}{x+2y} =\dfrac{3-1}{-1+2(1)}=2$