Why is the solution of the $\phi^6$ potential not a soliton? Consider a theory with a $\phi^6$-scalar potential:
$$
\mathcal{L} = \frac{1}{2}(\partial_\mu\phi)^2-\phi^2(\phi^2-1)^2.
$$
I solved its equation of motion but found that the general form of its solution,
$$
\phi(x) = \frac{e^x}{\sqrt{e^{2x}-C_1}}-\frac{e^{-x}}{\sqrt{e^{-2x}-C_2}},
$$
is not a soliton. Why isn't this solution to the equation of motion a soliton?
 A: Why do you say it is not a soliton solution?
I did not verify your answer but assuming its true, then it seems like a soliton to me.
Your vacuum consists of $\phi=0,\pm 1$. This solution obviously interpolates between the two vacua $\phi(+\infty)\rightarrow +1$ and $\phi(-\infty)\rightarrow -1$, and is therefore topologically stable. Moreover the region where potential energy is stored in this solution is localized. 
Try to play with $C_1$ and $C_2$ to visually see the location and dimensions of the kink.
A: Just to complement Ali Moh's answer:
We can define a topological current in the same way we do for the $\phi^4$ kink,
$$J_\mu=C\epsilon_{\mu\nu}\partial^\nu\phi(t,x),$$
where $C$ is a normalization constant and $\epsilon_{01}= -1$. The topological charge then is
$$Q=\int_{-\infty}^\infty J_t dx=C\int_{-\infty}^\infty\partial_x\phi dx=C\left[\phi(\infty,t)-\phi(-\infty,t)\right].$$
Using the solution given in the question (I am also assuming it is true) and setting $C=1/2$ we get
$$Q=1.$$
Notice that the fact the solution interpolates two distinct vacua is enough for a conserved topological charge. Inded, $\partial_\mu J^\mu=0$ implies
$$\frac{dQ}{dt}=\int_{-\infty}^\infty \partial_x J_x dx=J_x(\infty,t)-J_x(-\infty,t)=C\left[\frac{\partial\phi}{\partial t}(-\infty,t)-\frac{\partial\phi}{\partial t}(\infty,t)\right]=0,$$
since asymptotically $\phi\rightarrow\pm 1$ .
