Let $S$ be a regular surface and $X:U \subset \mathbb{R}^2\longrightarrow X(U)\subset S\subset \mathbb{R}^3$ a local parametrization. Does the following hold?
If $e_1, e_2$ are two linearly independent vectors in $\mathbb{R}^2$, and we assume $\mathbb{R}^2$ is embedded in $\mathbb{R}^3$ (we assume $X:\mathbb{R}^3\longrightarrow\mathbb{R}^3$)
¿$dX(e_1\times e_2)$ parallel to $dX(e_1)\times dX(e_2)$?
Note that the cross product $e_{1} \times e_{2}$ depends on whether you're interpreting these vectors as elements of the plane (in which case the product is customarily viewed as the scalar $1$) or as elements of space (in which case $e_{1} \times e_{2} = e_{3}$).
The first line of your question suggests $X$ is defined (only) on $U$, in which case $dX(e_{1} \times e_{2})$ doesn't make sense.
If instead you mean that $X$ is defined on some neighborhood of the plane in space (and your surface is obtained by restricting $X$ to an open set in the plane), then "no, $dX(e_{1} \times e_{2})$ is not generally parallel to $dX(e_{1}) \times dX(e_{2})$". Indeed, this isn't even true for linear transformations $X$. If we write $X$ as a $3 \times 3$ real matrix, then $$ dX(e_{1} \times e_{2}) = dX(e_{3}) $$ is the third column of $X$, while $dX(e_{1}) \times dX(e_{2})$ is the cross product of the first two columns of $X$.
There's no reason for the cross product of the first two columns to be proportional to the third column.