The function $\mathrm{f}:\left[-\frac{1}{2},\frac{1}{2}\right]\to \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ defined by $\mathrm{f}\left(\mathrm{x}\right)={\sin }^{-1}\left(3\mathrm{x}-4{\mathrm{x}}^3\right)$ is

Since ${\sin }^{-1}\left(3\mathrm{x}-4{\mathrm{x}}^3\right)=3{\sin }^{-1}\mathrm{x}\in \left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

$i.e. {\sin }^{-1}\mathrm{x}\in \left[-\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{6}\right]\text{ or }\mathrm{x}\in \left[-\frac{1}{2},\frac{1}{2}\right]$ .

$\boxed{\mathrm{So} \mathrm{f} \mathrm{is} \mathrm{onto}.}$ More ever the function $\mathrm{y}=\sin \mathrm{x}$ is one-one on $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$ so if $\mathrm{f}\left({\mathrm{x}}_1\right)=\mathrm{f}\left({\mathrm{x}}_2\right)$then
${\sin }^{-1}\left(3{\mathrm{x}}_1-4{\mathrm{x}}_1^3\right)={\sin }^{-1}\left(3{\mathrm{x}}_2-4{\mathrm{x}}_2^3\right)$ which implies that
$3{\mathrm{x}}_1-4{\mathrm{x}}_1^3=3{\mathrm{x}}_2-4{\mathrm{x}}_1^3={\sin }^{-1}\left(3{\mathrm{x}}_2-4{\mathrm{x}}_2^3\right)$which implies that $3{\mathrm{x}}_1-4{\mathrm{x}}_1^3=3{\mathrm{x}}_2-4{\mathrm{x}}_2^3\text{. }$The real solution of the last equation is given by ${\mathrm{x}}_1={\mathrm{x}}_2.$$\boxed{\mathrm{Hence} \mathrm{f} \mathrm{is} \mathrm{one}-\mathrm{one}}$