I was trying to find a counterexample to the theorem that states that principal curvature directions are orthogonal. Obviously such an example doesn't exist but I'm having hard time understanding what's wrong with the following example.
I construct the surface as follows. I take a planar curve, namely a parabola $a (r^2)$ and revolve it through the $Z$ axis while adjusting the parameter $a$ that controls the curvature of the parabola. The parametrization is given by:
$$ \left [X = rcos(\theta), \quad \quad Y = rsin(\theta), \quad \quad Z = ar^2 \right],$$ $$a(\theta)=0.5(1 + sin(4\theta)),$$
where the parametric domain is $$ \theta \in [0,\pi], \quad \quad r \in [-0.5,0.5].$$
The "problematic" point is the origin $(0,0,0)$. See a Matlab rendering of the surface in $3D$:
For $\theta \in \{3\pi/8,7\pi/8\}$ the $\sin(4\theta)$ in the expression for $a(\theta)$ is at minimum ($-1$) and we get $a=0$. The restriction of $\theta$ to these two values corresponds to two orthogonal straight lines on the surface (green in the figure). Hence $\kappa_2 = 0$. For $\theta \in \{\pi/8,5\pi/8\}$ the $\sin(4\theta)$ in the expression for $a(\theta)$ is at maximum ($1$) and we get $a=1$. The restriction of $\theta$ to these two values corresponds to two orthogonal parabolas on the surface (red in the figure) with $\kappa_1=2$.
Here is a top view of the $XY$ plane:
The "contradiction" seems to be that the curves with minimal and maximal curvatures at the origin are attained at $45$ degrees rather than $90$.