I am looking at the following exercise:
State an analogue of Theorem 1.5.1 for level curves in $\mathbb{R}^3$ given by $f(x, y, z) = g(x, y, z) = 0$.
The Theorem 1.5.1 is the following:
So, is the analogue for level curves in $\mathbb{R}^3$ as follows?
Let $f(x, y,z)$ and $g(x,y,z)$ be two smooth functions of three variables. Assume that, at every point of the level curve $$C = \{(x, y, z) \in \mathbb{R}^3 \mid f(x, y, z)=g(x,y,z)=0\}$$ $\nabla f$ and $\nabla g$ are not simultaneously zero. If $p$ is a point of $C$, with coordinates $(x_0, y_0, z_0)$, say, there is a regular parametrized curve $\gamma (t)$, defined on an open interval containing $0$, such that $\gamma$ passes through $p$ when $t = 0$ and $\gamma (t)$ is contained in $C$ for all $t$.
Or is it in this case somehow different?
In general, $C$ won't be a regular curve without imposing some extra conditions. For example, if $f = g$, then generically $C$ will be a surface. Even if $f = 0$ and $g = 0$ are two different surfaces, their intersection can be singular as demonstrated in the following image (taken from Wikipedia):
A popular condition that forces the intersection to be a regular curve is to require that the surfaces intersect transversely in $p$. The technical requirement is that $\nabla_f|_p$ and $\nabla_g|_p$ must be linearly independent. This means that the tangent planes to $f = 0$ and $g = 0$ at $p$ are not the same and so the surfaces are not tangent to each other at $p$. An inverse function theorem argument then shows that locally, around $p$, the intersection $C$ is the image of a regular curve.