#### Answer

$1.99$

#### Work Step by Step

Given that the function is a differentiable function with $f(x,y)=1-xycos(\pi y)$ at $1,1)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=1-xycos(\pi y)$
$f_{x}(a,b)=-ycos(\pi y)$
$f_{y}(a,b)=\pi xysin(\pi y)-xcos(\pi y)$
At $(1,1)$
$f(1,1)=1-(1)(1)cos(\pi (1))=1-(-1)=2$
$f_{x}(1,1)=-1cos(\pi (1))=1$
$f_{y}(1,1)=\pi (1)(1)sin(\pi (1))-1cos(\pi (1))=\pi(0)-1(-1)=1$
The linearization $L(x,y)$ of function at $(1,1)$ is
$L(x,y)= f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$
$=2+1(x-1)+1(y-1)$
$=2+x-1+y-1$
$L(x,y)=x+y$
Therefore, the linearization $L(x,y)$ of the function $f(x,y)=1-xycos(\pi y)$at $(1.02,0.97)$ is
$f(1.02,0.97)=1.02+0.97$
$=1.99$
The graph of function and tangent plane at (1, 1) is depicted below: