How would I go about showing that the vector field $X = x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}$ is tangent to the unit sphere in $\mathbb{R}^3$? I can see it pictorially but I'm not sure what would be required to actually prove it.
(This isn't homework! I understand the definitions of vector field as either a derivation of $C^\infty(M)$ or as a section of the tangent bundle $TM$, but I don't know how to compute anything about it so I'm trying to work through some examples.)
Here's how to make the argument from the comments a little more rigorous from basic properties of tangent bundles. Let $i: S^2 \to \mathbb R^3$ be the inclusion. For every $x \in S^2$ this induces an injection $i_* T_xS^2 \to T_x \mathbb R^3$. Then asking if a vector field $X$ on $\mathbb R^3$ is tangent to $S^2$ is the same as asking if for all $x \in S^2, X_x \in i_* T_x S^2$. Now consider the map $f: \mathbb R^3 \to \mathbb R, (x,y,z) \mapsto x^2 + y^2 + z^2$. Now $f \circ i : S^2 \to \mathbb R$ is constant so that $f_* \circ i_* = 0$. In other words, $i_* TS^2 \subset \ker f_*$. Since $f_* \ne 0$ on $S^2$, by dimensionality reasons we must have that $i_* TS^2 = \ker f_*$. Finally, a vector field is in $\ker f_*$ is equivalent to $Xf = 0$ since $Xf = df(X) = f_* X$.
The unit sphere in $\Bbb R^3$ is nice in the sense that its tangent spaces admit a simple interpretation: If $r \in S^2$, then $$T_r S^2 = \{v \in \Bbb R^3 : r \cdot v = 0 \}.$$ Now we can make the identification $$x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \leftrightarrow (-y, x, 0) \in \Bbb R^3.$$ Now for any $r = (x,y,z) \in S^2$, we have $$(x, y,z) \cdot (-y, x,0) = -xy + xy + 0= 0 \implies (-y, x, 0) \in T_r S^2.$$ This holds for all points $r \in S^2$, so $x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x}$ is a vector field on $S^2$.
Since we're explicitly considering an embedding, you can use the concept of a vector manifold to attack this problem. There are many benefits to doing so: in particular, you can use geometric (clifford) algebra to analyze the problem.
What Henry has done is clever: he's using the normal to the surface to characterize it, which is something you can always do in $\mathbb R^3$, but not something you can do generally. For instance, if your manifold $M$ has $\dim M = 2$ in $\mathbb R^4$, you can't construct a vector normal to it.
Geometric algebra allows you to characterize an orientable manifold by its pseudoscalar. This is what I will do here.
The unit sphere can be characterized by two angles $\theta, \phi$. We assign each point a vector in $\mathbb R^3$, which is obvious by the embedding. Let these vectors be $r$ (so that $r^2$, the second component of this vector, is $y$ as usual). We can then take, at any $r$, partial derivatives with respect to our angular coordinates to characterize tangent vectors.
$$\begin{align*}e_\theta &= \frac{\partial x}{\partial \theta} = \cos \theta \cos \phi e_1+ \cos \theta \sin \phi e_2 - \sin \theta e_3 \\ e_\phi &= \frac{\partial x}{\partial \phi} = \sin \theta (-\sin \phi e_1 + \cos \phi e_2)\end{align*}$$
For this problem, it helps to rewrite the variables back in terms of $x, y, z$.
$$\begin{align*}e_\theta &= \frac{zx}{\sqrt{x^2 + y^2}} e_x+ \frac{zy}{\sqrt{x^2+y^2}}e_y - \sqrt{x^2+y^2} e_z \\ e_\phi &= - y e_x + x e_y\end{align*}$$
We can now characterize the surface by its pseudoscalar $i$, which is formed by a wedge product of the tangent vectors.
$$i = e_\theta \wedge e_\phi = \sqrt{x^2 + y^2} (z e_{xy} + y e_{zx} + x e_{xy})$$
The pseudoscalar represents, for a surface, an area element. It is a function of position, as you can see. Because it has orientation, the pseudoscalar itself can tell you about the orientation of a manifold (indeed, if you were to get that the pseuscalar is multivalued, you would rightly conclude that the manifold is non-orientable). The pseudoscalar is used to perform projections under the geometric product. If a vector field lies entirely in the tangent space, then its wedge product with the pseudoscalar should be zero. We can see that this is indeed the case.
$$i \wedge X \propto (z e_{xy} + y e_{zx} + x e_{yz}) \wedge (x e_y - y e_x) = yx e_{zxy} - xy e_{yzx} = (yx -xy) e_{xyz} = 0$$
This approach is a bit more complicated for $\mathbb R^3$ than the standard approach, and the clifford algebra aspects of it may take some getting used to, but it's a powerful formalism when you have access to an embedding, and we could've just as easily used it for a 2-dimensional surface in $\mathbb R^4$ with almost no change in approach.
