Answer
$x+y+z=0$
Work Step by Step
$z=x+sin(x+y)$, $(-1,1,0)$
Consider $f(x,y)= x+sin(x+y)$
$f_{x}(x,y)=sin(x+y)+xcos(x+y)$
$f_{y}(x,y)=xcos(x+y)$
At $(-1,1,0)$
$f_{x}(-1,1)=sin(-1+1)+1.cos(-1+1)=-1$
$f_{y}(x,y)=-1cos(-1+1)=-1$
The equation of the tangent plane to the given surface at the specified point $(-1,1,0)$ is given by
$z-z_{0}=f_{x}(x_{0},y_{0})(x-x_{0})+f_{y}(x_{0},y_{0})(y-y_{0})$
On substituting the values, we get
$z-0=-1(x+1)-1(y-1)$
$z=-x-1-y+1$
Hence, $x+y+z=0$
