Interpreting $\nabla {g}$ when $g(u,x,y)=0$. Say I have $u=f(x,y)=x^2+y^2$. Then $u-x^2-y^2=0$. We can write $g(u,x,y)=u-x^2-y^2=0$. 
As a result, we have $\nabla {g}=(1,-2x,-2y)$. 
How do we interpret $\nabla {g}$. If we were to plot $g$ for various values of $x$ and $y$, we'd get $0$ everywhere. But $\nabla {g}$, which could be interpreted as the rate of change of $g$ is non-zero at many points. For example, $g(2,1,1)=0$. Here, $\nabla {g}=(1,-2,-2)$. 
Thanks in advance!
 A: You are confusing yourself with poor notation.
If you have $f(x,y) = x^2+y^2$, this defines a real valued function defined everywhere. It takes values in $[0,\infty)$.
If you pick some $u \in [0,\infty)$, and consider the set of $(x,y)$ pairs that satisfy the equation $f(x,y) = u$, that is, $L_u = \{ (x,y) | f(x,y) = u \}$ then you are no longer considering arbitrary pairs of $(x,y)$ pairs, but only considering those that satisfy the equation $f(x,y) = u$.
By reusing $f$ as in $f(u,x,y) = 0$ you are confusing things and the notation police will show up at your door. Let us use $\phi(u,x,y) =u-f(x,y)$ instead.
Now note that $\phi(u,x,y) = 0 $ iff $(x,y) \in L_u$. This does not mean that $\phi$ is zero everywhere. For example, if $\phi(u,x,y) = 0$, then we have $\phi(u+1,x,y) = 1$. The equation $\phi(u,x,y) = 0$ defines a surface $G \subset \mathbb{R}^3$ (in fact, it is the graph of the function $f$).
If you pick a point $(u',x',y') \in G$, then $\nabla \phi(u',x',y')$ is a normal to the surface $G$ at the point $(u',x',y')$.
A: You start with the $(x,y)$-plane and there have a function $f(x,y):=x^2+y^2$. Then you look at the graph 
$$G_f:=\{(x,y,u)\>|\> (x,y)\in{\mathbb R}^2, \ u=f(x,y)\}$$ of this function, which is a surface in three-dimensional $(x,y,u)$-space. This surface can be viewed as solution set of the equation $u-x^2-y^2=0$. The left hand side
$g(x,y,u):= x^2+y^2-u$ of this equation is a function of three variables, and our graph $G_f$ is a level surface of this function $g$. It follows that at each point $(x,y,u)\in G_f$ the gradient $\nabla g(x,y,u)=(-2x,-2y,1)$ is orthogonal to the tangent plane of $G_f$ at $(x,y,u)$. If we project this gradient onto the $(x,y)$-plane we obtain $\nabla'g=(-2x,-2y)=-\nabla f(x,y)$. The latter "plane" gradient vector is at each point $(x,y)\in{\mathbb R}^2$ orthogonal to the level line of $f$ through $(x,y)$.
