$\Omega\subset\subset\mathbb{R}^N$ strongly "analytically convex" at a point $p\in\partial\Omega$ implies that all curvatures at $p$ are positive I'm reading Convex Analysis by Krantz, and I'm stuck on a remark he makes about "analytic convexity" (which is apparently equivalent to convexity), which he defines as follows:

Let $\Omega\subseteq\mathbb{R}^n$ be a bounded domain with $C^2$ boundary and $\rho$ a defining function for $\Omega$. Fix a point $p\in\partial\Omega$. We say that $\partial\Omega$ is analytically (weakly) convex at $p$ if
  $$
\sum_{j,k=1}^{N}\frac{\partial^2\rho}{\partial x_j\partial x_k}(p)w_jw_k\geq 0,\;\;\;\;\;\;\;\;\;\;\forall w\in T_p(\partial\Omega).
$$
  We say that $\partial\Omega$ is strongly convex at $p$ if the inequality is strict whenever $w\neq 0$.

Reading further he mentions that

The reader may wish to verify that at a strongly convex boundary point, all curvatures are positive (in fact, one may, by the positive definiteness of the matrix $(\partial^2\rho/\partial x_j\partial x_k)$, impose a change of coordinates at $p$ so that the boundary of $\Omega$ agrees with a ball up to and including second order a $p$).

It turns out that I am a reader who wishes to verify this fact. (I'm less interested in the remark in parentheses, but I've included it for the sake of completion.) However I've failed to do this. I've been attempting (and failing) to prove the following case of it all day:

Let $\Omega\subset\subset\mathbb{R}^3$ be an open region with $C^2$ boundary. If $p\in\partial\Omega$ is a strongly convex boundary point (with $\Omega$ having $C^2$ boundary), then the Gaussian curvature of $\partial\Omega$ at $p$, $K(p)$, is positive.

Attempt
Well $\partial\Omega=\rho^{-1}(0)$, so we can invoke the implicit function theorem to get a local parametrization $(u,v)\mapsto (u,v,h(u,v))$, where $h:\mathbb{R}^2\to\mathbb{R}$ is such that $\rho(u,v,h(u,v))=0$. Now (omitting the calculations),
$$
K(u,v)=\frac{h_{uu}h_{vv}-h_{uv}^2}{(1+h_u^2+h_y^2)^2},
$$
and I can relate the derivatives of $\rho$ to the derivatives of $h$ via differentiating both sides of $\rho(u,v,h(u,v))=0$, but after a mess of computations I've failed to get any form which allows me to invoke the definition.
Is there a more simple way to approach this? Am I missing something obvious? I know this can be done using the more geometric definition of convexity, but I'd prefer to use this definition if at all possible. Help of any form is appreciated. Thanks in advance.
 A: I think got it. Thanks @studiosus for being patient with me. Given an arbitrary defining form $\rho:\mathbb{R}^3\to\mathbb{R}$, we can use the implicit function theorem to find a function $h:\mathbb{R}^2\to\mathbb{R}$ such that $\rho\big(x,y,h(x,y)\big)=0$. It follows (as above) that
$$
K(x,y)=\frac{h_{xx}h_{yy}-h_{xy}^2}{(1+h_x^2+h_y^2)^2}.
$$
Now  either $(x,y,z)\mapsto h(x,y)-z$ or $(x,y,z)\mapsto z-h(x,y)$ is a (local) defining form for $\Omega$.
Suppose it is the former. Call it $\tilde\rho$. Since the definition of analytic convexity is invariant of choice of defining form, the hessian of $\tilde\rho$ is positive-definite. Since
$$
\begin{pmatrix}\tilde\rho_{xx} & \tilde\rho_{xy}\\\tilde\rho_{xy} & \tilde\rho_{yy}\end{pmatrix}=\begin{pmatrix}h_{xx} & h_{xy} \\ h_{xy} & h_{yy}\end{pmatrix}
$$
is the $2$nd leading principal minor of the hessian of $\tilde\rho$, it has positive determinant; that is, $h_{xx}h_{yy}-h_{xy}^2$ is positive. This is the numerator of the expression for $K(x,y)$. Since the denominator for the expression is always positive, we conclude that $K(x,y)>0$.
If it is the latter, the same procedure applies (you get a negative infront of every entry of the $2$nd leading block matrix but this doesn't affect its determinant).
Thus $K(x,y)>0$.
