= Solution
Set
$$
U(x)=\frac{x^2}{(1+x^2)^2},
\qquad
E(x,y)=\frac12y^2+U(x).
$$
The stated identity gives
$$
U'(x)=\frac{2x(1-x^2)}{(1+x^2)^3}.
$$
Hence, along the system,
$$
\begin{aligned}
\dot E
&=y\dot y+U'(x)\dot x\\
&=y[-U'(x)-ky]+U'(x)y\\
&=-ky^2\leq0.
\end{aligned}
$$
Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne0$. The <First Lyapunov theorem> therefore gives stability.
Choose a sufficiently small compact <invariant sublevel set>
$$
\Omega_c=\{(x,y):E(x,y)\leq c\},
\qquad 0<c<\frac14.
$$
On $\{\dot E=0\}$ we have $y=0$. A trajectory can remain there only if
$$
\dot y=-U'(x)=0,
$$
which occurs at $x=0,\pm1$. The choice $c<1/4=U(\pm1)$ excludes the two latter points, so the largest invariant subset of $\{\dot E=0\}\cap\Omega_c$ is the origin. <LaSalle invariance principle> now gives convergence to the origin. Thus, for every $k>0$,
$$
\boxed{(0,0)\text{ is asymptotically stable}.}
$$
This is an instance of <damped mechanical energy as a Lyapunov function>.
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