So first of all, I know that the Weingarten map (which from now on I shall denote by $L$) is a symmetric linear operator, so there is an orthonormal basis of eigenvalues (Spectral Theorem).
I have been trying this concrete example for a while but I am just stuck and I would appreciate an extra pairs of eyes.
I will use the result that claims $L=G^{-1}H$ where $G$ is the matrix for the first fundamental form and $H$ is the matrix for the second fundamental form.
Our surface is given by $f(u,v)=(u\cos v, u\sin v , v)$. Then:
$f_u=(\cos v, \sin v, 0)$, $f_v=(-u\sin v, u \cos v, 1)$, $f_{uu}=(0,0,0)$, $f_{uv}=f_{vu}=(-\sin v, \cos v, 0)$, and $f_{vv}=(-u\cos v, -u \sin v, 0)$.
Thus, $g_{11}=f_u\cdot f_u=1$, $g_{12}=g_{21}=f_u\cdot f_v=0$, and $g_{22}=f_v\cdot f_v =u^2+1$.
Then, $f_u\times f_v=(\sin v, -\cos v, u)$, so $n=\frac{1}{\sqrt{1+u^2}}(\sin v, -\cos v, u)$.
Then, $h_{11}=n\cdot f_{uu}=0$, $h_{12}=h_{21}=n\cdot f_{uv}=\frac{-1}{\sqrt{u^2+1}}$, and $h_{22}=0$.
Then as $G^{_1}=Diag[1,\frac{1}{1+u^2}]$, we have that $L=G^{-1}H$, is: $L_{11}=0$, $L_{12}=\frac{-1}{\sqrt{u^2+1}}$, $L_{21}=\frac{-1}{(u^2+1)^{3/2}}$, and $L_{22}=0$ (here is where I start to doubt because I thought I would always wind up with a symmetric matrix).
I get that the eigenvalues of this matrix are given by $\det(\lambda I-L)=\lambda^2-\frac{1}{(1+u^2)^2}$, so $k_1=\frac{1}{u^2+1}$, and $k_2=-k_1$. Then to find the first principal direction, we have to find the eigenvector correspoding to $k_1$, which turned out to be $(\frac{-1}{\sqrt{u^2+1}},1)$, and the other principal direction turned out to be $(\frac{1}{\sqrt{u^2+1}},1)$, which are not always orthogonal, so I dont know where I went wrong.