Equilibrium points of the given system are $x_{eq} = 1,2$. The Jacobian is $J(x) = -[2(x-1)(x-2) + (x-2)^2]$. Evaluating the Jacobian at $x = 1, J(1) = -1$ and at $x = 2, J(2) = 0$. $\therefore$ the linearized system is $\dot{x}_2 = J(1)x_2 = -x_2$ about the point $x = 1$ and $\dot{x}_3 = J(2)x_3 = J(2)x_3 = 0$(Note: $x_2$ and $x_3$ are as defined in the question). The point $x_{eq} = 1$ is stable because $J(1)$ is negative while nothing can be commented on $x_{eq} = 0$ because $J(2) = 0$.
The system equation in terms of $x_2$ is $\dot{x_2} = -x_2(x_2 - 1)^2$ and in terms of $x_3$ is $\dot{x_3} = -(x_3 + 1)x_3^2$. Consider $ V_2(x_2) = (1/2)x_2^2, \dot{V}_2 = -x_2^2(x_2 - 1)^2 \leq 0$. Can't figure out from here onwards.