We use

$$

\begin{array}{l}

e^{x}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \quad \Rightarrow e^{-x^{2}}=1-x^{2}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\cdots \\

\sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots

\end{array}

$$

so that

$$

\begin{aligned}

e^{-x^{2}} \sin x &=\left(1-x^{2}+\frac{x^{4}}{2 !}-\frac{x^{6}}{3 !}+\cdots\right)\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right) \\

&=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots-x^{2}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right)+\frac{x^{4}}{2 !}\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots\right) \\

&=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots-x^{3}+\frac{x^{5}}{3 !}+\frac{x^{5}}{2 !}+\cdots \\

&=x-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots

\end{aligned}

$$

Hence

$$

\lim _{x \rightarrow 0} \frac{f(x)-x}{x^{3}}=\lim _{x \rightarrow 0} \frac{x-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots-x}{x^{3}}

$$

$$

\begin{array}{l}

=\lim _{x \rightarrow 0} \frac{-\frac{7 x^{3}}{6}+\frac{27 x^{5}}{40}+\cdots}{x^{3}} \\

=\lim _{x \rightarrow 0}\left(-\frac{7}{6}+\frac{27 x^{2}}{40}+\cdots\right) \\

=-\frac{7}{6}

\end{array}

$$