Find direction of normal and the equation of the tangent plane at the point.
z=sqrt(4-x^2-y^2)
at (1,1,sqrt(2))
I prefer to think of such a surface as being a level surface for g(x,y,z)= f(x,y)- z= 0 and using the fact that the gradient, $\displaystyle \nabla g= f_x\vec{i}+ f_y\vec{j}- \vec{k}$, is normal to a level surface. Of course, that gives exactly the same thing FernandoRevilla is saying.