How does Implicit Function Theorem guarantee curves through a surface?

So when proving that the gradient vector of some function $g(x,y,z)$ is always orthogonal to a level set of $g$, my book uses differentiable curves which lie on this level set. As far as I understand, these curves are guaranteed to exist by the Implicit Function Theorem, but I'm having trouble understanding why.

For a level set in $\textbf{R}^2$ I understand, because the implicit function theorem says we can solve for $y$ as a function of $x$ or the other way around. So we can get a function $y=f(x)$ and then we can just define the curve $\varphi(t) = \langle t, f(t) \rangle$. But how does the same conclusion follow for a level set in $\textbf{R}^3$?

• It goes through the same: you can solve for one of the three variables, say $z$, in terms of the other two. However many equations you have, you lose that many independent variables (in the generic situation). – Ian Oct 9 '15 at 3:04
• I guess the reason I'm confused is because I thought if you have a parametric equation with two variables, then it is a parametric surface, not a curve. So like if you solved for z as a function of x and y then made a parametric equation out of that. – Fgilan Oct 9 '15 at 3:10
• Yes, you get a parametric surface, not a curve, from a single scalar equation in three variables. You can then look at curves on this surface if you want. – Ian Oct 9 '15 at 3:23

The implicit function theorem shows that there is a local map $z : \Bbb R^2 \to \Bbb R$ such that $f(x, y, z(x, y))$ is constant. Any curve in the domain of $z$ in $\Bbb R^2$ is carried by the map $(x, y) \mapsto (x, y, z(x, y))$ to a curve in the level surface.