I have a past qual question here: given a smooth embedding $f \colon S^2 \to \mathbb{R}^3$, show that there must exist distinct points $p,q \in S^2$ such that the tangent planes to the embedded sphere $f(S^2)$ at $f(p)$ and $f(q)$ are parallel.
My thought was that, assuming the tangent planes to $f(p)$ and $f(q)$ are not parallel for all $p$ and $q$, to construct a vector field on $f(S^2)$ such that its pullback by $f$ is nonvanishing, which would yield a contradiction since we have an even-dimensional sphere.
Any hints or suggestions are appreciated!
I think there is something simpler.
Let $x,y,z$ denote the coordinates in $\mathbb{R}^3$. By compactness, there is $p\in f(S^2)$ such that $z(p)$ is the maximal value of $z$ along $f(S^2).$ Any path $\alpha:(-\epsilon,\epsilon)\to f(S^2),\;t\mapsto(x(t),y(t),z(t))$ with $\alpha(0)=p$, satisfies $\dot{z}(0)=0$. It follows that $T_pf(S^2)\subset\{(a,b,0)|a,b\in\mathbb{R}\}$, and for dimension reasons, $T_pf(S^2)=\{(a,b,0)|a,b\in\mathbb{R}\}$. Now take $q\in f(S^2)$ with minimal $z$, and use a similar argument to conclude that $T_pf(S^2)=T_qf(S^2)$.
Note that we only use compactness.