Answer
$\begin{aligned} f_{x}(0,-1,-2) &=1 \\ f_{y}(0,-1,-2) &=2 \\ f_{z}(0,-1,-2) &=3 \\ f(0,-1,-2) &=-4 \end{aligned}$
Work Step by Step
Given that $f(x, y)$ is differentiable at (0,-1,2) and partial derivatives exist, we use the Local Linear Approximation Theorem:
$f\left(x_{0}, y_{0}, z_{0}\right)+f_{f}\left(x_{0}, y_{0}, z_{0}\right)\left(x-x_{0}\right)+f_{f}\left(x_{0}, y_{0}, z_{0}\right)\left(y-y_{0}\right)+f_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(z-z_{0}\right)=L(x, y, z)$
So:
$\Delta x=-x_{0}+x, \Delta y=-y_{0}+y$ , $\Delta z=-z_{0}+z$, $x_{0}=0, y_{0}=-1, and\ z_{0}=-2$
Thus:
$x+2 y+3 y+4=f(0,-1,-2)+f_{x}(0,-1,-2)(x-0)+f_{y}(0,-1,-2)(y-(-1))$
$+f_{z}(0,-1,-2)(z-(-2))$
$1=f_{x}(0,-1,-2)$
$2=f_{y}(0,-1,-2)$
$3=f_{z}(0,-1,-2)$
Solving for $f(0,-1,-2)$
$f(0,-1,-2)=4-(0) f_{2}(0,-1,-2)-(1) \int_{y}(0,-1,-2)-(2) 5_{z}(0,-1,-2)$
$4-(1) 2-(2)(3)=f(0,-1,-2)$
$-4=f(0,-1,-2)$
