Answer
$2$
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{y-3}{x+2y} \times (-1)=\dfrac{-y+3}{x+2y} $
Now, plug the point $(-1,1)$
we get $\dfrac{-y+3}{x+2y} =\dfrac{-1+3}{-1+2(1)}=2$
