How to pick a Lyapunov function and prove stability?

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.