Answer
$\triangle z=-0.7189$ and $dz=-0.73$
Work Step by Step
Write the differential form $dz=\dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy + \dfrac{\partial z}{\partial z} dz$
The partial derivatives w.r.t $t$ and $x$ are:
$z_x=2x-y$ and $z_y=6y-x$
At the given points, we have $z_x(3,-1)=7$ and $z_y(3,-1)=-9$
Here, $\triangle x=2.96-3=-0.04\\ \triangle y=-0.95-(-1)=0.05$
Further,
$dz=\dfrac{\partial z}{\partial x} dx + \dfrac{\partial z}{\partial y} dy + \dfrac{\partial z}{\partial z} dz=7 \times (-0.04) +(-9) \times (0.05)=-0.73$
and
$\triangle z=f(3,-1)-f(2.96,-0.95)=14.2811-15=-0.7189$