The Lyapunov function is
$$
V(u,u')= \int_0^u g(u)\, du+\frac12 (u')^2.
$$
Indeed, $g(0)=0$, $g'(0)>0$ implies that for sufficiently small $u>0$, $g(u)$ is positive, and for sufficiently small $u<0$, $g(u)<0$, thus, in some neighborhood of $0$,
$\int_0^u g(u)\, du>0$. Hence, $V(u,u')$ is positive definite. Obviously, $V(0,0)=0$ and $V$ is continuously differentiable.
The derivative
$$
V'= \frac{\partial V}{\partial u} u'+\frac{\partial V}{\partial u'}u''=
g(u)u'+u'(-g(u))=0,
$$
thus, the origin is stable. Since $V(u,u')$ is the first integral of the system, it is not asymptotically stable.
The periodicity of the solutions is following from the fact that the level sets of Lyapunov functions are diffeomorphic to the sphere (see, for instance, this article).
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.
Best Answer
Lyapunov theorems can be proved using only the assumption of continuity, where one says that a Lyapunov function is a continuous function $V$ which decreases (weakly) along every orbit (see for example Zehnder's Lectures on Dynamical Systems, Section IV.2). The thing is that the usual way of checking that a function is decreasing is by using the derivative, so for simplicity most sources add the assumption that $V$ is differentiable and that $\dot V \le 0$.
Regarding the phrase continuously differentiable, it does not mean “continuous and differentiable” (which would indeed be redundant), but that the function can be differentiated and that the result of the differentiation is continuous. The ending “-ly” turns the adjective continuous into an adverb, indicating that the function is “differentiable in a continuous way”. To be explicit: we say that a multivariate function $V(x_1,\dots,x_n)$ is continuously differentiable if all the partial derivatives $\partial V/\partial x_i(x_1,\dots,x_n)$ exist and are continuous. (This is a simple sufficient condition for $V$ to be differentiable in the multivariate sense.)