For $k=1,2,\dots$, let $$I_k:=\int_{|x-k\pi|<1/k}x^4\exp(-2x^8\sin^2 x)\,dx.$$ Then, as $k\to\infty$, $$I_k\asymp k^4\int_{|x-k\pi|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\ =k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8\sin^2 x\}\,dx \\ =k^4\int_{|x|<1/k}\exp\{-(2+o(1))(k\pi)^8 x^2\}\,dx \asymp1,$$ by substitution $u=(k\pi)^4 x$. Hence, $$\int_{-\infty}^\infty f(x)^2\,dx\ge\sum_{k=1}^\infty I_k=\infty.$$ So, $f$ is actually not square-integrable.
Reasoning similarly (but using, say, $h_k:=1/k^{b/3}$ instead of $1/k$ in $|x-k\pi|<1/k$), one can see that for any real $a,b>0$, letting $$f(x):=|x|^a\exp(-|x|^b\sin^2x),$$ we have the following:
$f$ is continuous, but unbounded at $\infty$.
$f$ is square-integrable iff $2a-b/2<-1$.
To get this result, we also note that for $k=1,2,\dots$ $$\int_{h_k\le|x-k\pi|\le\pi}x^{2a}\exp(-2x^b\sin^2 x)\,dx \\ =O(k^{2a}\exp\{-(2+o(1))(k\pi)^b h_k^2\})=O(1/k^c)$$ for any real $c$.
In your example, we have $a=2$ and $b=8$, so that $2a-b/2=0\not<-1$, and so, your $f$ is not square-integrable.