Is the form $dx_1 \wedge dx_2$ vanishing somewhere? Consider the form $d x_1 \land dx_2$ on $\mathbb{R}^2$. Does it vanish anyhere? My calculation shows that for tangent vectors $v_p, w_p$ at a point $p \in \mathbb{R}^2$: $$(dx \land dy)(p) = dx\vert_p(v_p) dy\vert_p(w_p) - dx\vert_p(w_p)dy\vert_p(v_p)$$ Is this calculation correct? Would it be correct to say that this implies that $(dx \land dy)(p)$ is not the zero-map for all $p \in \mathbb{R}^2$? Hence, $dx \land dy$ is non-vanishing and therefore a volume form on $\mathbb{R}^2$. Any input is appreciated.

For $v_p=(v_1,v_2)$ and $w_p=(w_1,w_2)$ you have that

$$\begin{align}(dx\wedge dy)(p)(v_p,w_p)&=dx(p)(v_p)dy(p)(w_p)-dx(p)(w_p)dy(p)(w_p)\\\ &=v_{1}w_2-w_1v_2\\\ &=\det\begin{pmatrix}v_1&w_1\\\v_2&w_2\end{pmatrix}\end{align}$$

this is equal to zero if and only if the vectors $v_p$ and $w_p$ are linearly dependent. Since on the tangent plane at $p$ you always have a pair of tangent vector that are not linearly dependent, then $dx\wedge dy$ is not the zero form at every point $p$.