#### Answer

$2x+2y+(\pi-4)$

#### Work Step by Step

Given that f is a differentiable function with $f(x,y)=4arctan(xy)$ at $(1,1)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=4arctan(xy)$
$f_{x}(a,b)=\frac{4y}{(1+x^{2}y^{2})}$
$f_{y}(a,b)=\frac{4x}{(1+x^{2}y^{2})}$
At $(1,1)$
$f(1,1)=4arctan(1)=4.\frac{\pi}{4}=\pi$
$f_{x}(a,b)=\frac{4.1}{(1+1^{2}1^{2})}=2$
$f_{y}(a,b)=\frac{4.1}{(1+1^{2}1^{2})}=2$
The linearization $L(x,y)$ of function at $(1,1)$ is
$L(x,y)= f(1,1)+f_{x}(1,1)(x-1)+f_{y}(1,1)(y-1)$
$=\pi+2(x-1)+2(y-1)$
$=\pi+2x-2+2y-2$
$=2x+2y+(\pi-4)$
Hence, the linearization $L(x,y)$ of the function at $(1,3)$ is $2x+2y+(\pi-4)$