The question I am attempting reads as follows:
Let $\theta$ and $\phi$ be standard spherical coordinates on $S^2 = \{x^2 + y^2 + z^2 =1\}$. Recall that $\theta, \phi$ are defined as
$$x=\cos\phi \cos\theta,$$ $$y=\sin\phi \cos\theta,$$ $$ z=\sin\theta,$$
and can be considered as local coordinates on the whole of $S^2$ except for the north and south poles. Let $P=(\sqrt{2},\sqrt{2},0) \in S^2$ and $\xi = (1,-1,1)$. Check that $\xi$ (as a vector from $\mathbb{R}^3$) is tangent to $S^2$ at $P$. Find the coordinates of $\xi$ in the spherical coordinates $(\phi, \theta).$
First of all I assume that this point $P$ should have some kind of factor involved such that its modulus is equal to 1 - otherwise it wouldn't lie on the unit sphere. So for calculations I guess we take $P$ to actually be given by $P = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0).$ From here I don't really know hw to approach this. For context, I am doing a postgraduate course in Lie groups and Lie algebras and this falls into the introduction of some introductory differential geometry. The definition of a tangent to a manifold we have been given is that it is a tangent vector if it is a derivation at $P$. If you could prove it using this definition that would be great. I would also appreciate it if you could spare a proof using another method if there exists a more intuitive way.
$\xi$ is a tangent vector to the unit sphere at $P$ if and only if $\xi \cdot \vec{OP}=0.$ Since $(1,-1,1)\cdot \left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right)= 0$ we can say that $\xi=(1,-1,1)$ is a tangent vector to the unit sphere at the point $P\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},0\right).$
Now, how to write $\xi=(1,-1,1)$ in spherical coordinates? We have $$\sqrt{3}(\cos\phi \cos\theta,\sin\phi \cos\theta,\sin\theta)=(1,-1,1).$$ If we consider third coordinates we have $\sin\theta=\dfrac{1}{\sqrt 3}\implies \theta =\arcsin \dfrac{1}{\sqrt 3}.$ Now, from
$$\sqrt 3\cos\phi \cos \arcsin \dfrac{1}{\sqrt 3}=1$$ we get $$\phi=\arccos \dfrac{1}{\sqrt 3 \cos \arcsin \dfrac{1}{\sqrt 3}}.$$