Based on Thomas Ferguson's Linear Programming: A Concise Introduction, I would solve it like this:
First, construct the simplex tableau.
\begin{align*}
\begin{array}{c | rrr | r}
& x_1 & x_2 & x_3\\
\hline
y_1 & -1 & -1 & -1 & -2\\
y_2 & 2 & -1 & 1 & 1\\
\hline
& -2 & 6 & 0 & 0
\end{array}
\end{align*}
Using case 2 ($b_i\leq 0$) from p. 24, pivot on entry $(2,1)$.
\begin{align*}
\begin{array}{c | rrr | r}
& y_2 &x_2 & x_3\\
\hline
y_1 & \frac{1}{2}& -\frac{3}{2} & -\frac{1}{2} & -\frac{3}{2}\\
x_1 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & \frac{1}{2}\\
\hline
& 1 & 5 & 1 & 1
\end{array}
\end{align*}
Then, using case 1 from the minimizing problem ($-\mathbf{c}\geq \mathbf{0}$) p. 26, pivot on entry $(1,2)$.
\begin{align*}
\begin{array}{c | rrr | r}
& y_2 &y_1 & x_3\\
\hline
x_2 & -\frac{1}{3}& -\frac{2}{3} & \frac{1}{3} & 1\\
x_1 & \frac{1}{3} & -\frac{1}{3} & \frac{2}{3} & 1\\
\hline
& \frac{8}{3} & \frac{10}{3} & -\frac{2}{3} & -4
\end{array}
\end{align*}
Finally, we use case 1 ($\mathbf{b}\geq\mathbf{0}$) from p. 23 and pivot on entry $(2,3)$.
\begin{align*}
\begin{array}{c | rrr | r}
& y_2 &y_1 & x_1\\
\hline
x_2 & -\frac{1}{2}& -\frac{1}{2} & -\frac{1}{2} & \frac{1}{2}\\
x_3 & \frac{1}{2} & -\frac{1}{2} & \frac{3}{2} & \frac{3}{2}\\
\hline
& 3 & 3 & 1 & -3
\end{array}
\end{align*}
Since both $\mathbf{c}$ and $\mathbf{b}$ are positive, the solution can be read off as $x_2=\frac{1}{2}$, $x_3=\frac{3}{2}$ and the rest of the variables zero. The maximum value $-3$ is in the lower right corner.
EDIT: After looking at the given link, I'll add a solution using the slack-variables for future reference.
Adding the slack-variables, we get the following problem
\begin{align*}
\text{Max }z=2x_1-6x_2,\quad&\text{s.t.}\\
-x_1-x_2-x_3+w_1&=-2\\
2x_1-x_2+x_3+w_2&=1\\
x_i,w_j&\geq 0.
\end{align*}
First, a feasible solution must be found. Since the right-hand side is negative, we cannot simply choose $x_i=0$, since this would contradict $w_1\geq 0$. Instead, it may be seen that letting $x_1=x_3=w_1=0$ and thus $x_2=2$ and $w_2=3$ is a feasible solution to the problem. Therefore, we obtain the system
\begin{align*}
x_2 &= 2+w_1-x_1-x_3\\
w_2 &= 3-3x_1+w_1-2x_3\\
z &= -12+ 8x_1-6w_1+6x_3.
\end{align*}
Looking at $z$, it may be increased by adding either $x_1$ or $x_3$ to the solution. Normally we would add $x_1$ since this will have the greatest effect on the value of $z$, but since I already know that $x_1=0$ in the optimal solution, I choose $x_3$ (otherwise, we would remove $x_1$ from the solution in a later step). This gives a new system by isolating $x_3$ in the second equation and substituting in the expression for $x_2$.
\begin{align*}
x_3 &= \frac{3}{2} -\frac{3}{2}x_1+\frac{1}{2}w_1-\frac{1}{2}w_2\\
x_2 &= \frac{1}{2} + \frac{1}{2}w_1+\frac{1}{2}x_1+\frac{1}{2}w_2\\
z &= -3 -x_1-3w_1-3w_2.
\end{align*}
Since all the coefficients in $z$ are negative, adding $x_1, w_1$ or $w_2$ to the solution would lower the value. Thus, we set $x_1=w_1=w_2=0$ and obtain the solution $x_2=\frac{1}{2}$ and $x_3=\frac{3}{2}$. The optimal value is $z=-3$.
