Answer
$$\frac{1}{6}$$
Work Step by Step
Since
\begin{align*}
\lim _{y \rightarrow 0} \frac{\sin ^{-1} y-y}{y^{3}}
&=\frac{0}{0}
\end{align*}
Then by using L’Hôpital’s Rule
\begin{align*}
\lim _{y \rightarrow 0} \frac{\sin ^{-1} y-y}{y^{3}}&=\lim _{y \rightarrow 0} \frac{\frac{1}{\sqrt{1-y^2}} -1}{3y^{2}}\\
&=\lim _{y\to \:0}\left(\frac{1-\sqrt{1-y^2}}{3y^2\sqrt{1-y^2}}\right)\\
&= \lim _{y\to \:0}\left(\frac{\frac{y}{\sqrt{-y^2+1}}}{\frac{3\left(-3y^3+2y\right)}{\sqrt{1-y^2}}}\right)\\
&= \lim _{y\to \:0}\left(\frac{1}{3\left(-3y^2+2\right)}\right)\\
&=\frac{1}{6}
\end{align*}
