#### Answer

$2x+y-1$

#### Work Step by Step

Given that f is a differentiable function with $f(x,y)=x^{2}e^{y}$ at $(1,0)$
The linearization $L(x,y)$ of function at $(a,b)$ is given by
$L(x,y)= f(a,b)+f_{x}(a,b)(x-a)+f_{y}(a,b)(y-b)$
$f(x,y)=x^{2}e^{y}$
$f_{x}(a,b)=2xe^{y}$
$f_{y}(a,b)=x^{2}e^{y}$
At $(1,0)$
$f(1,0)=1^{2}e^{0}=1$
$f_{x}(1,0)=2(1)e^{0}=2$
$f_{y}(1,0)=1^{2}e^{0}=1$
The linearization $L(x,y)$ of function at $(1,0)$ is
$L(x,y)= f(1,0)+f_{x}(1,0)(x-1)+f_{y}(1,0)(y-0)$
$=1+2(x-1)+1(y-0)$
$=1+2x-2+y$
$=2x+y-1$
Hence, the linearization $L(x,y)$ of the function at $(1,4)$ is $ 2x+y-1$