When are the eigenvalues of the second fundamental form equal to the principal curvatures? I am confused about the following concerning the second fundamental form.
Consider a surface $S$ $\subset R^3$
If we consider a chart at a point $p \in S$, $f$: $R^2$$\to S$  and suppose $\partial f/\partial x$ and $\partial f/\partial y$ are orthonormal, Does it then follow that the eigenvalues of the second fundamental form are the principal curvatures and the eigenvectors are the principal directions?
Moreover If  $\partial f/\partial x$ and $\partial f/\partial y$ are not orthonormal this may not be true?
Thanks in advance
 A: Let $S$ be a regular surface, $f$ a local parametrization in some neighborhood of a point $p$, and $f_{x}$, $f_{y}$ the coordinate vector fields, and $N$ a continuous unit normal field. Notation below follows the exam mentioned in the comments.
The first fundamental form is the quadratic form on $T_{p}S$ whose matrix with respect to the basis $\{f_{x}, f_{y}\}$ is
$$
I = \left[\begin{array}{@{}cc@{}}
    f_{x} \cdot f_{x} & f_{x} \cdot f_{y} \\
    f_{x} \cdot f_{y} & f_{y} \cdot f_{y} \\
  \end{array}\right]
 = \left[\begin{array}{@{}cc@{}}
    g_{11} & g_{12} \\
    g_{12} & g_{22} \\
  \end{array}\right].
$$
The second fundamental form is the quadratic form on $T_{p}S$ whose matrix with respect to the basis $\{f_{x}, f_{y}\}$ is
$$
II = \left[\begin{array}{@{}cc@{}}
    N \cdot f_{xx} & N \cdot f_{xy} \\
    N \cdot f_{xy} & N \cdot f_{yy} \\
  \end{array}\right]
  = \left[\begin{array}{@{}cc@{}}
    e & f \\
    f & g \\
  \end{array}\right].
$$
The shape operator is the endomorphism associated to $II$ by "raising an index" with respect to $I$. Its matrix in the basis $\{f_{x}, f_{y}\}$ is
$$
\frac{1}{g_{11} g_{22} - g_{12}^{2}}\left[\begin{array}{@{}rr@{}}
     g_{22} & -g_{12} \\
    -g_{12} &  g_{11} \\
  \end{array}\right]\left[\begin{array}{@{}cc@{}}
    e & f \\
    f & g \\
  \end{array}\right],
$$
its eigenvalues are the principal curvatures, and its eigenvectors are the principal directions.
If the coordinate basis $\{f_{x}, f_{y}\}$ is orthonormal at $p$, then the matrix of the first fundamental form at $p$ is the identity, so the matrix of the shape operator coincides with the matrix of the second fundamental form, just as you've noticed; if the coordinate basis is not orthonormal, however, the matrices of the second fundamental form and the shape operator do not have the same entries.
