Answer
$\dfrac{4}{3}$
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{3x^2+y}{-4y+x} \times (-1)=\dfrac{-3x^2-y}{-4y+x} $
Now, plug the point $(1,1)$
we get $\dfrac{-3x^2-y}{-4y+x} =\dfrac{-3-1}{-4+1}=\dfrac{4}{3}$