how to solve system of linear equations of XOR operation?

As I wrote in my comment, you can just use any method you know for solving linear systems, I will use Gauss:

$$ \begin{array}{cccc|c||l} \hline x & y & z & w &\ & \\ \hline\hline 1 & 1 & 1 & 0 & 1 & \\ 1 & 1 & 0 & 1 & 1 & \text{$+$ I}\\ 1 & 0 & 1 & 1 & 0 & \text{$+$ I}\\ 0 & 1 & 1 & 1 & 1 & \\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \text{III}\\ 0 & 1 & 0 & 1 & 1 & \text{II}\\ 0 & 1 & 1 & 1 & 1 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 1 & 0 & 0 & \text{$+$ III}\\ \hline 1 & 1 & 1 & 0 & 1 & \\ 0 & 1 & 0 & 1 & 1 & \\ 0 & 0 & 1 & 1 & 0 & \\ 0 & 0 & 0 & 1 & 0 & \\\hline \end{array} $$ Now we can conclude $w = 0$ from line 4, which gives $z = 0$ from 3 and $y = 1$ from 2, and finally $x = 0$. So $(x,y,z,w) = (0,1,0,0)$ is the only solution.

I'll try it the 'artisanal way'.

These four rules should be enough : $\begin{align} &x\oplus 0=x\\ &x\oplus 1=\overline{x}\\ &x\oplus x=0\\ &x\oplus \overline{x}=1\\ \end{align}$

$\tag{1}1=x\oplus y\oplus z$
$\tag{2}1=x\oplus y\oplus w$
$\tag{3}0=x\oplus w\oplus z$
$\tag{4}1=w\oplus y\oplus z$

Applying $x\oplus$ on the three first equations and $w\oplus$ on the last gives :

$\tag{5}\overline{x}=y\oplus z$
$\tag{6}\overline{x}=y\oplus w$
$\tag{7}x= w\oplus z$
$\tag{8}\overline{w}=y\oplus z$

$(6)\oplus(7)\ $ gives $\ \tag{6'}1=y\oplus z$ so that from $(8)$ $w=0$
from $(7)$ and $(6)$ $\ x=z$ and $\ \overline{x}=y$
from $(5)$ $\ \overline{x}=1$ and the final conclusion :
$$(w,x,y,z)=(0,0,1,0)$$ as found (earlier) by martini...