I have a vector v=$\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}$ which I will rotate using Euler rotation. After I tested some vectors and some rotations I found out that to reverse a Euler rotation you have invert the rotation order and then rotate by the negative amount of rotation in each axis.
So if I have a vector v=$\begin{pmatrix} v_1\\v_2\\v_3 \end{pmatrix}$ which I first rotate by $r_1$ around the x-axis then by $r_2$ around the y-axis then $r_3$ around the z-axis. To get my original vector v again I have to first rotate by $-r_3$ around the z-axis then by $-r_2$ around the y-axis then $-r_1$ around the x-axis.
I tried to search for proofs of this online however everything I found was proved using matrices, which I unfortunately am not acquainted with.
How can you prove that to reverse a Euler rotation you have to invert the rotation order and then rotate by the negative amount of rotation in each axis without using matrices?
This is just a general property of functions (notice that your rotations are just functions acting in your vectors), namely that the inverse of $f\circ g$ is $g^{-1}\circ f^{-1}$. If this notation is not familiar to you, it means that if
$$y=f(g(x)),$$
then
$$x=g^{-1}(f^{-1}(y)).$$
To see why, simply notice that
$$g^{-1}(f^{-1}(y))=g^{-1}(f^{-1}(f(g(x))))=g^{-1}(g(x))=x,$$
which can also be written as
\begin{align*} (g^{-1}\circ f^{-1})\circ(f\circ g)&=g^{-1}\circ(f^{-1}\circ f)\circ g\\ &=g^{-1}\circ\mathrm{Id}\circ g \\ &=g^{-1}\circ g\\ &=\mathrm{Id}, \end{align*}
and similarly for the other direction. Note that this also of course requires invertibility of $f$ and $g$, and in your case, with rotations, clearly they are invertible (you can always just revert a rotation by rotating back).