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!

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 $(u',x',y')$.
