Answer
\[
\begin{aligned}
1=f_{x}(3,2,1) & \\
-1=f_{y}(3,2,1) & \\
2=f_{z}(3,2,1) & \\
1=f(3,2,1) &
\end{aligned}
\]
Work Step by Step
Given that $f(x, y)$ is differentiable at (3,2,1) and partial derivatives exist, we use the Local Linear Approximation Theorem:
$\left.\left.L(x, y, z)=f\left(x_{0}, y_{0}, z_{0}\right)+f_{z}\left(x_{0}, y_{0}, z_{0}\right)\right)\left(x-x_{0}\right)+\int_{y}\left(x_{0}, y_{0}, z_{0}\right)\right)\left(y-y_{0}\right)+\int_{z}\left(x_{0}, y_{0}, z_{0}\right)\left(z-z_{0}\right)$
$\Delta x=-x_{0}+x, \Delta y=-y_{0}+y$, $\Delta z=-z_{0}+z$, $3=x_{0}, 2=y_{0}, and \ 1=z_{0}$
Thus:
$x-y+2 z-2=f(3,2,1)+f_{x}(3,2,1)(-3+x)+f_{y}(3,2,1)(-2+y)+f_{z}(3,2,1)(-1+z)$
$f_{x}(3,2,1)=1$
$f_{y}(3.2 .1)=-1$
$f_{z}(3,2,1)=2$
Therefore, we have
$-2+(3) f_{\{}(3,2,1)+(2) f_{y}(3,2,1)+(1) z_{t}(3,2,1)=f(3,2,1)$
$-2+(3)(1)+(2)(-1)+(1)(2)=f(3,2,1)$
$1=f(3,2,1)$
