Explanation
At equilibrium position
$$\displaystyle \mathrm{F}=-\dfrac{dV}{dx}=0$$$$\dfrac{dV}{dx }=\dfrac{C(a^{2}-x ^{2})}{(x ^{2}+a^{2})^{2}}=0$$$$\therefore $$ There are two equilibrium positions$$x_1 = a$$$$x_2 = -a$$Consider $$\dfrac{d^{2}V}{dx ^{2}}=\dfrac{2Cx (x ^{2}-3a^{2})}{(x^{2}+a^{2})^{3}}$$$$\dfrac{d^{2}V}{dx ^{2}}|_{x_1}< 0$$$$\dfrac{d^{2}V}{dx ^{2}}|_{x_2}> 0$$
$$\because $$There is a maxima at $$x=a$$ and minima at $$x=-a$$$$\Rightarrow x _{1}$$ is a position of unstable equilibrium and $$x _{2}$$ is a position of stable equilibrium.
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