Proving regularity for this ratio I have a (strongly convex) hypersurface $S\subseteq\mathbb R^n$ and two defining functions $f,g:U\to\mathbb R$, where $U$ is an open neighborhood of a point $x\in S$, i.e. $U\cap S=\{f=0\}=\{g=0\}$. I therefore know the ratio $\frac fg$ (and its reciprocal) are defined, and have at least the lowest of the regularities of $f$ and $g$, throughout the whole of $U\smallsetminus S$. But how do I show I can extend that ratio to the whole $U$ (i.e. define it on $S\cap U$) in such a way as to preserve such regularity?
 A: As a matter of fact, we can only guarantee slightly lower regularity of the ratio. Explicitly, if both defining functions are $C^k$, the following shows that their ratio is $C^{k-1}$. 
Assume both defining functions are $C^k$. Since $S$ is a regular surface, we may change the coordinates if necessary and assume $$S=\{x_n=0\}.$$Furthermore, none of the points on $S$ is a singular point of $f$, and it follows that on $S$ we have $f_n\neq0,$ where $f_i$ stands for $\partial f/\partial x_i$. The same holds for $g$, of course.
Lemma: There is a non-vanishing $C^{k-1}$ function $h:U\to\mathbb{R}$, such that $f=x_nh$.
Proof: Define $F:U\times[0,1]\to\mathbb{R}$ by$$F(x_1,\ldots,x_n,u)=f(x_1,\ldots,x_{n-1},ux_n).$$By the chain rule, we have $F_u=x_nf_n$. Hence,
\begin{align}
f(x_1,\ldots,x_n)&=f(x_1,\ldots,x_n)-f(x_1,\ldots,x_{n-1},0)\\
&=F(x_1,\ldots,x_n,1)-F(x_1,\ldots,x_n,0)\\
&=\int_0^1F_u(x_1,\ldots,x_n,u)du\\
&=x_n\int_0^1f_n(x_1,\ldots,x_{n-1},ux_n)du.
\end{align}
So, we define $$h(x_1,\ldots,x_n)=\int_0^1f_n(x_1,\ldots,x_{n-1},ux_n)du,$$and $h$ is clearly $C^{k-1}$. Also, $h$ does not vanish far away from $S$ since $f$ doesn't. The fact that $h$ does not vanish on $S$ follows from $f_n\neq0$ on $S$. QED Lemma
Similarly, there is a non-vanishing $C^{k-1}$ function $h'$ such that $g=x_nh'$. The ratio $h/h'$ is well defined everywhere, and we're done.
