As $x\to \infty$, $\sin (x^2)$ is bounded between $-1$ and $1$, and $\sin (1/x^2)\sim 1/x^2$. Also ${\log }_e(1+x)\sim {\log }_{e}x$ for large $x$. So the expression behaves like

$\frac{(\text{bounded})\cdot ({\log }_{e}x{)}^{\alpha }\cdot \frac{1}{x^2}}{x^{\alpha \beta }({\log }_{e}x{)}^{\beta }}=\frac{({\log }_{e}x{)}^{\alpha -\beta }}{x^{\alpha \beta +2}}\cdot (\text{bounded}).$

For the limit to be $0$, the dominant factor $x^{\alpha \beta +2}$ in the denominator must grow without bound, which requires $\alpha \beta +2>0$, i.e. $\alpha \beta >-2$.

Checking each option:

(A) $\alpha \beta =-3$. Fails.

(B) $\alpha \beta =-1>-2$. Holds.

(C) $\alpha \beta =-1>-2$. Holds.

(D) $\alpha \beta =-2$. Boundary case, fails the strict inequality.

$\therefore$ Correct options are (B) and (C).