Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 833: 10

$z=1$

The vector equation is given by: $r(x,y,z)=r_0+t \nabla f(r_0)$ Given: $e^{-x^2-y^2}-z=0$ The equation of tangent line for $\nabla f(0,0,1)=\lt 0,0,-1 \gt$ is given a follows: $(0)(x-0)+(0)(y-0)-(1)(z-1)=0 \implies -z+1=0$ Thus, $z=1$