Answer
a) $df=2 dx +1 dy$; The function $f$ is more sensitive to a change in $x$.
b) $\dfrac{-1}{2}$
Work Step by Step
a) $f_x=2x (y+1) \\ \implies f_x(1,0)=2x(y+1)=2(1)(0+1)=2$
Next, $f_y=x^2 \\ \implies f_y(1, 0)=$
So, $df=2 dx +1 dy$
We can see that the function $f$ is more sensitive to a change in $x$.
b) $df=0$
$2 dx +1 dy=0$
This gives: $\dfrac{dx}{dy}=\dfrac{-1}{2}$
