Answer
$x+1$
Work Step by Step
Given $f(x,y)=e^xcos(xy)$
Linear approximation : $L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$
$z_0=e^0cos(0)=1; f_x=e^xcos(xy)-ye^xsin(x)\\f_y=-xe^xsin(xy)$
Then, we have $f_x(0,0)=1; f_y(0,0)=0$
Therefore,
$L(x,y)=L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0=1(x-0)+0(y-0)+1=x+1$
