Suppose that $r:\mathbb{R}^2\to\mathbb{R}^3$ is $C^1$ and the cross product $$\frac{\partial r}{\partial u}\times\frac{\partial r}{\partial v}(u_0,v_0)$$ is nonzero for some $(u_0,v_0)\in\mathbb{R}^2.$ Show that $r$ is the graph of a $C^1$ function $\mathbb{R}^2\to\mathbb{R}$ on a neighborhood of $(u_0,v_0).$
I believe I should be using the Implicit Function Theorem, but I'm getting tripped up on the details. For instance, I tried letting $$s:\mathbb{R}^3\times\mathbb{R}^2\to\mathbb{R}^3,$$ $$s(x,(u,v)) = x-r(u,v),$$ which isn't what we need.
I just need some hints please, not complete answers!
Ok, no sulution, but some hints, as requested...
$r_u\times r_v\neq 0 $ means $r_u$ and $r_v$ are nonzero and linear independent. By continuity, this is true in a neighbourhood of $(u_0,v_0)$. In other words, $r$ defines locally a two dimensional manifold which is known to admit a local representation as a graph.
If this unknown to you choose a basis vector $e_i$ such that $r_u(u_0,v_0), r_v(u_0,v_0), e_i$ are linear independent. Wlog $e_i= e_3$ corresponding to the $z$ coordinate. Now look at $$F:\mathbb{R}^2\rightarrow \mathbb{R}^2$$ defined by $$(u,v)\mapsto \langle r,e_2\rangle e_2+ \langle r,e_1\rangle e_1$$
This (is just the projection of the image of $r$ to the $(x,y)$ plane and) has to be a map with a differential of maximal rank from what has been discussed above, so you can apply the implicit function theorem. So $F$ is a local diffeomorphism near $(u_0,v_0)$ and so the projection $(x,y,z)\mapsto (x,y) $, when restricted to the image of $r$ near $(u_0,v_0)$ is one-to-one and onto.