Answer
$x+y-1$
Work Step by Step
Given $f(x,y)=)=\dfrac{y-1}{x+1}$
Linear approximation : $L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0$
$z_0=\dfrac{0-1}{0+1}=-1; f_x=-\dfrac{y-1}{(x+1)^2} \\f_y=\dfrac{1}{x+1}$
Then, we have $f_x(0,0)=1; f_y(0,0)=0$
Therefore,
$L(x,y)=L(x,y)=f_x (x-x_0) + f_y (y-y_0)+z_0=1(x-0)+1(y-0)+(-1)=x+y-1$