For a surface $\mathbf{r} = \mathbf{r}(s, t)$ are the partial derivatives $\mathbf{r}_s$ and $\mathbf{r}_t$ in general orthogonal?
I was thinking the surface of a sphere and in that case indeed these vectors are orthogonal at every point. We can think the partial derivatives as the tangent vectors of the curves $\mathbf{r} (s, t_0)$ and $\mathbf{r} (s_0, t)$ respectively. Can the above fact for the surface of a sphere be generalized?
Your example is apparently $$\mathbf r(\theta, \phi) = (\cos \theta\sin\phi, \sin\theta\sin\phi,\cos\phi)$$ where it is indeed true that $\mathbf r_\theta \cdot \mathbf r_\phi = 0$. But notice that this a standard coordinate parametrization of the sphere. This coordinate system was chosen exactly for the property that the $\theta$ and $\phi$ coordinate directions are perpendicular to each other.
In general, this is not true. Another familiar way of parametrizing half the sphere is $$\mathbf r(x, y) = (x, y,\sqrt{1 - x^2 - y^2})$$ for which we find $$\mathbf r_x = (1, 0, \frac {-x}{\sqrt{1 - x^2 - y^2}})\\ \mathbf r_y = (0, 1, \frac {-y}{\sqrt{1 - x^2 - y^2}})$$ and thus $$\mathbf r_x \cdot \mathbf r_y = \frac {xy}{1 - x^2 - y^2}$$ which is only $0$ when either $x$ or $y$ is $0$.