I'm currently reading about sections in tangent bundles, and there's very simple example
The collection of vector fields $\partial/\partial x, \partial/\partial y, \partial/\partial z$ is a smooth frame on $\mathbb{R}^3$.
I think specifically it is meant $\left. \partial/\partial x \right|_p,\left. \partial/\partial y \right|_p,\left. \partial/\partial z \right|_p$ I was trying to prove this using the provided definitions,
Definition A frame of a vector bundle $\pi : E \to M$ over an open set $U$ is a collection of sections $s_1,\ldots,s_r$ of $E$ over $U$ such that at each point $p \in U$, the elements $s_1(p),\ldots,s_r(p)$ form a basis for the fiber $E_p := \pi^{-1}(p)$. A frame $s_1,\ldots,s_r$ is said to be smooth or $C^{\infty}$ if $s_1,\ldots,s_r$ are $C^{\infty}$ sections of $E$ over $U$.
I think it's easy enough to show that $\left. \partial/\partial x \right|_p,\left. \partial/\partial y \right|_p,\left. \partial/\partial z \right|_p$ are sections of $T \mathbb{R}^3$, to show that for fixed $p$ they form a basis I define the vector
$$ X_p = \alpha \left. \partial/\partial x \right|_p + \beta \left. \partial/\partial y \right|_p +\gamma \left. \partial/\partial z \right|_p = 0 $$
And applying $X_p$ to the functions $x,y$ and $z$ gives me the linear independence. For the smoothness I'm not really sure. I think the reason is because the example doesn't really define the vector bundle precisely in this context. I think though I can take $E = T\mathbb{R}^3, M = \mathbb{R}^3$ and $\pi$ as the projection map I get a vector bundle, but this means that for each $p \in \mathbb{R}^3$ I have $\pi^{-1}(p) = \left\{ p \right\} \times \mathbb{R}^3$, and I can give to $\pi^{-1}(p)$ a vector space structure of dimension $0$, how do I get the continuity from here?
You want the $\partial/\partial x$ not just $\partial/\partial x|_p$, which is a vector in the tangent space at the point $p$. The $\partial/\partial x$ are vector fields, that is one vector in the tangent space at each point. In more fancy language, these are sections of the bundle $E=T\mathbb{R}^3$ over $M=\mathbb{R}^3$. Note that $E$ is isomorphic to $\mathbb{R}^6$. So $\pi$ is not the identity map but rather the projection to the base point. So $\pi^{-1}(p)=T_pM$ which is isomorphic to $\mathbb{R}^3$. The $\partial/\partial x$ are smooth sections because if you identify each tangent space with $\mathbb{R}^3$ there are just constant.