The map $\varphi:(\theta,\phi)\rightarrow\Big(f(\theta,\phi),\theta,\phi\Big)$ is not a parametric representation of the spherical surface $r=f(\theta,\phi)$. Rather, $\varphi$ is a parametric representation of the surface $x=f(y,z)$ since parametric surfaces (and curves) are inherently represented in Cartesian form. What you're doing is equivalent to saying $\theta \longrightarrow \Big(\theta^2,\theta\Big)$ is a parametric represention of the polar spiral $r=\theta^2$ when, in reality, $\theta \rightarrow \Big(\theta^2,\theta\Big)$ is parametric representation of the parabola $x=y^2$ while $\theta \rightarrow \Big(\theta^2 \cos(\theta),\theta^2 \sin(\theta)\Big)$ is the spiral $r=\theta^2$.
Let's refer to your example in which you seek to compute the surface area of a sphere.
In spherical coordinates, the equation of a sphere is $r=1$ on the domain $(\theta,\phi)\in[0,2\pi)\times [0,\pi]$. You can represent this parametrically as $$(\phi,\theta) \longrightarrow \Big(\sin(\phi)\cos(\theta),\sin(\phi)\sin(\theta),\cos(\phi)\Big)$$ simply by converting from spherical to cartesian coordinates. However, the image of your function $\varphi:(\phi,\theta)\longrightarrow (1,\phi,\theta)$ on the same domain is the rectangular patch $\{1\}\times [0,2\pi)\times [0,\pi]$ embedded in the vertical plane $x=1$ which has area $2\pi^2$. This is precisely why $$\int_0^{2\pi}\int_0^{\pi}\sqrt{1+\big(f_{\theta}\big)^2+\big(f_{\phi}\big)^2}d\phi d\theta=2\pi^2 \neq 4\pi$$
whenever $f(\phi,\theta)=1$; this integral is calculating the area of the surface $x=1$ on $(y,z)\in [0,2\pi)\times[0,\pi]$ which is the image of your map $\varphi$.
If you want to find the area of the surface given in spherical coordinates by $r=f(\phi,\theta)$ defined on domain $(\phi,\theta)\in \mathcal{U}$ you will need to express this parametrically as $$\vec{p}(\phi,\theta)=\Big(f(\phi,\theta)\sin(\phi)\cos(\theta),f(\phi,\theta)\sin(\phi)\sin(\theta),f(\phi,\theta)\cos(\phi)\Big)$$ The area of the surface will be $$S=\int \int _{\mathcal{U}}||\vec{p}_\phi \times \vec{p}_\theta||d\phi d\theta$$ If you compute $||\vec{p}_\phi \times \vec{p}_\theta||$ you will surely obtain your desired result.
You want to rotate the surface so that the direction $(1,0,0)$ becomes $\dfrac1{\sqrt3}(1,1,1)$. A possible rotation matrix is
$$\begin{pmatrix}\frac1{\sqrt3}&0&-\frac2{\sqrt6}
\\\frac1{\sqrt3}&\frac1{\sqrt2}&\frac1{\sqrt6}
\\\frac1{\sqrt3}&-\frac1{\sqrt2}&\frac1{\sqrt6}\end{pmatrix}$$
Now your equation is
$$\frac{(u+v+w)^2}3-\frac{(v-w)^2}2-\frac{(-2u+v+w)^2}6=1$$
or
$$u^2+v^2+w^2-4uv-4vw-4wu+3=0.$$
Caution: sign and permutations errors are guaranteed, I didn't check. But the final result must not be far from correct.
https://www.wolframalpha.com/input/?i=plot+x%5E2%2By%5E2%2Bz%5E2-4xy-4yz-4zx%2B3%3D0
Best Answer
HINTS: First you should check that the points given by the parametrization do in fact lie on the hyperboloid. Next, you want to see that the $u$- and $v$-curves are in fact lines. You need to do some algebraic manipulations with the rational functions (like "long division"). It will help to fix $v=v_0$, say.