Eigenvectors give only the direction of principal axes (for central conics and quadrics) not the principal curvature directions. There's a paper for finding Gaussian and mean curvatures for implicit surfaces. The standard results for general quadrics are summarized as follows:
Using matrix representation,
\begin{align}
0 &= F(x,y,z) \\
& \equiv \frac{1}{2}
\begin{pmatrix}
x & y & z & 1 \\
\end{pmatrix}
\begin{pmatrix}
a & h & g & i \\
h & b & f & j \\
g & f & c & k \\
i & j & k & d
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z \\ 1
\end{pmatrix} \\
\\
\nabla F &=
\begin{pmatrix}
a & h & g & i \\
h & b & f & j \\
g & f & c & k
\end{pmatrix}
\begin{pmatrix}
x \\ y \\ z \\ 1
\end{pmatrix} \\
\\
K &= -\frac{1}{|\nabla F|^4} \det
\begin{pmatrix}
a & h & g & i \\
h & b & f & j \\
g & f & c & k \\
i & j & k & d
\end{pmatrix} \\
\\
H &=
\frac{(\nabla F)^T
\begin{pmatrix}
a & h & g \\
h & b & f \\
g & f & c
\end{pmatrix}
(\nabla F)}{2|\nabla F|^3}-
\frac{a+b+c}{2|\nabla F|} \\
\\
\kappa_{\pm} &= H \pm \sqrt{H^2-K}
\end{align}
The principal directions or lines of curvature are not easy to solve, please refer to this paper for your further interest.
For standard paraboloids, the lines of curvature are as follows:
- Elliptic paraboloid $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=\dfrac{2z}{c}$
$$\boldsymbol{x}=
\begin{pmatrix}
\frac{a}{c} \sqrt{a^2-b^2} \sinh u \cos v \\
\frac{b}{c} \sqrt{a^2-b^2} \cosh u \sin v \\
\frac{a^2-b^2}{4c} (\cosh 2u-\cosh 2v)
\end{pmatrix}$$
- Hyperbolic paraboloid $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=\dfrac{2z}{c}$
$$\boldsymbol{x}=
\begin{pmatrix}
\frac{a}{c} \sqrt{a^2+b^2} \sinh u \cosh v \\
\frac{b}{c} \sqrt{a^2+b^2} \cosh u \sinh v \\
\frac{a^2+b^2}{4c} (\cosh 2u-\cosh 2v)
\end{pmatrix}$$
Analogous to quadratic polynomials:
\begin{align}
\mathbb{A}^{T} &= \mathbb{A} \\
f(\boldsymbol{x}) &=
\boldsymbol{x}^{T} \mathbb{A} \, \boldsymbol{x}+
\boldsymbol{b}^{T} \boldsymbol{x}+c \\
&=
(\boldsymbol{x}-\boldsymbol{h})^{T} \mathbb{A}
(\boldsymbol{x}-\boldsymbol{h})+k \\
\boldsymbol{h}
&=
-\frac{1}{2} \, \mathbb{A}^{-1} \boldsymbol{b} \\
k &=
c-\frac{1}{4} \, \boldsymbol{b}^{T} \mathbb{A}^{-1} \, \boldsymbol{b} \\
\nabla f(\boldsymbol{x}) &= 2\mathbb{A} \, \boldsymbol{x}+\boldsymbol{b} \\
\nabla^{2} f(\boldsymbol{x}) &= 2\operatorname{tr}(\mathbb{A})
\end{align}
Note that $\boldsymbol{h}$ is the centre of central quadric (or conic) providing $\det \mathbb{A} \ne 0$.
For paraboloids, $$\det \mathbb{A}=0$$
See also another answer of mine for using Hessian matrix to find the curvature of superellipse.
