I think a direct argument just based on the direction of the vector field should work.
Let me write $(x,y)$ instead of $(x_1,x_2)$, for simplicity. Then the ODEs are $\dot x = y-x^3$ and $\dot y = -y^3$. So it's clear that all trajectories have the property that $y(t) \to 0$ monotonically as $t \to \infty$.
Now consider to begin with the region $x>0$ and $0<y<x^3$, i.e., the region between the positive $x$-axis (which is a $y$-nullcline) and the right half of the curve $y=x^3$ (which is an $x$-nullcline). In that region, $\dot x$ and $\dot y$ are both negative, so a trajectory $(x(t),y(t))$ starting in that region cannot exit it (since it can't cross those nullclines; the vector field isn't pointing out of the region anywhere on those curves). So $x(t)$ is bounded from below (by zero) and decreasing, and must therefore have a limit $x^* \ge 0$ as $t \to \infty$. Thus, $(x(t),y(t)) \to (x^*,0)$, and the only possibility is then that $x^*=0$, since the points $(x^*,0)$ with $x^*>0$ aren't equilibria.
Next, if you start in the upper half plane, but to the left of the curve $y=x^3$, you have $\dot x>0$ and $\dot y<0$, so you will go downwards to the right. Then either $(x(t),y(t)) \to (0,0)$ directly, monotonically in both $x$ and $y$ (which I don't think actually can happen, but that doesn't matter), or you will eventually cross the curve $y=x^3$ and enter into the region covered in the first case, which will force you to tend to the origin from there.
Similarly if you start anywhere in the lower half plane, because of the rotational symmetry of the system. And if you start on the $x$-axis, it's also clear that you tend to the origin.
So all trajectories tend to the origin.
And the origin is a stable equilibrium to begin with; this wasn't immediately clear to me, but I convinced myself as follows: given a neighbourhood $U$, one can take a rectangle $V = (-a,a) \times (-a^3,a^3)$ with $a>0$ small enough so that $V \subseteq U$, and then it can be seen by similar arguments as above that trajectories starting in $V$ will stay in $V$, and hence in $U$.
Best Answer
The only equilibrium point of $$dy/dt = 1 − e^y,\; −∞ < y_0 < ∞.$$ is at $dy/dt =0.$
Thus equilibrium happens at $$ e^y =1 $$ that is $y=0$
This equilibrium is asymptotically stable because $y>0 \implies dy/dx <0$ and $ y<0 \implies dy/dx>0.$
Therefore the solutions approach the equilibrium solution $y=0$ asymptotically.