To my knowledge there is not a general method for finding a Lyapunov function. In this case one can solve the differential equations and use that to find a Lyapunov function. Namely $x_2$ is decoupled from $x_1$ and can be shown to have the following solution
$$
x_2(t) = C_1\,e^{-t},
$$
where $C_1$ is a constant and depends on the initial condition of $x_2$. Substituting the above equation into the expression for $\dot{x}_1$ gives
$$
\dot{x}_1 = x_1 (C_1\,e^{-t} -1)
$$
which is a separable differential equation, namely
$$
\frac{dx_1}{x_1} = (C_1\,e^{-t} -1) dt.
$$
Integrating on both sides gives
$$
\log(x_1) = -C_1\,e^{-t} -t+C_2.
$$
Solving for $x_1$ gives
\begin{align}
x_1(t) &= e^{-C_1\,e^{-t} -t+C_2}, \\ &= C_3\,e^{-C_1\,e^{-t} -t}, \\
&= C_3\,e^{-t}\,e^{-C_1\,e^{-t}},
\end{align}
or when using the definition for $x_2$ then it can also be expressed as $x_1(t)=C_3\,e^{-t}\,e^{-x_2}$. So the quantities $x_2$ and $x_1\,e^{x_2}$ will both decay exponentially fast, so the following Lyapunov function can be used
$$
V(x) = x_2^2 + x_1^2\,e^{2\,x_2},
$$
for which it can be shown that its derivative is
$$
\dot{V}(x) = -2\,x_2^2 - 2\,x_1^2\,e^{2\,x_2}.
$$
I will leave proving that $V(x)$ is radially unbounded to you.
"Global" and "asymptotical" are different attributes. Note that a stable equilibrium may not be an attractor. For instance, the solutions of:
$$
x'=-y \\
y'=x
$$
are the circles $x^2+y^2=r_0^2$ so the origin is stable because trajectories remain inside a bounded neighbourhood as small as desired (depending on the initial $r_0$), but they do not tend to the origin.
Since this system is linear, the stability is global, meaning there are no unbounded trajectories even if we start far from the origin. In contrast, for the one-dimensional system:
$$
y'=y(1-y)
$$
linearization shows that the equilibrium $y^*=1$ is stable and, in fact, asymptotically stable. However, this property is not global, e.g. the solution with initial condition $y_0=0$ is the constant trajectory $y(t)\equiv 0$ so $y(t)\notin B_{\frac{1}{2}}(1)$.
Best Answer
You have the system
$$ \begin{align} \dot{x}_1&=-x_1-x_2\\ \dot{x}_2&=-x_1-x_2^3 \end{align} $$
The linearization matrix of this system is
$$ \begin{pmatrix}-1&-1\\-1&-3x_2^2\end{pmatrix} $$
so at the origin we have
$$ \begin{pmatrix}-1&-1\\-1&0\end{pmatrix} $$
which has eigenvalues $-\frac{1}{2} \pm \frac{\sqrt{5}}{2}$, or approximately, $0.6180$ and $-1.6180$. Since both eigenvalues are different from zero and have a different sign (one strictly positive, one strictly negative), the origin is unstable.
Btw., note that the system cannot be globally stable because you have multiple (three) equilibrium points: $(0,0),(1,-1),(-1,1)$ all make $\dot{x}=0$, so even if the linearization would be stable, the system could not be globally asymptotically stable.