Answer
If $y=f(u)$ and $u=g(x)$
By using The Chain Rule we have:
$$\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}$$
so,
$$\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left(\frac{d y}{d x}\right)\\
&=\frac{d}{d x}\left(\frac{d y}{d u} \frac{d u}{d x}\right)\\
& \,\,\,\,\,\,\,\,\,\,\ \text{ [Product Rule] }\\
&=\left[\frac{d}{d x}\left(\frac{d y}{d u}\right)\right] \frac{d u}{d x}+\frac{d y}{d u} \frac{d}{d x}\left(\frac{d u}{d x}\right) \\
&=\left[\frac{d}{d u}\left(\frac{d y}{d u}\right) \frac{d u}{d x}\right] \frac{d u}{d x}+\frac{d y}{d u} \frac{d^{2} u}{d x^{2}}\\
&=\frac{d^{2} y}{d u^{2}}\left(\frac{d u}{d x}\right)^{2}+\frac{d y}{d u} \frac{d^{2} u}{d x^{2}}
\end{aligned}$$
Work Step by Step
If $y=f(u)$ and $u=g(x)$
By using The Chain Rule we have:
$$\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}$$
so,
$$\begin{aligned}
\frac{d^{2} y}{d x^{2}} &=\frac{d}{d x}\left(\frac{d y}{d x}\right)\\
&=\frac{d}{d x}\left(\frac{d y}{d u} \frac{d u}{d x}\right)\\
& \,\,\,\,\,\,\,\,\,\,\ \text{ [Product Rule] }\\
&=\left[\frac{d}{d x}\left(\frac{d y}{d u}\right)\right] \frac{d u}{d x}+\frac{d y}{d u} \frac{d}{d x}\left(\frac{d u}{d x}\right) \\
&=\left[\frac{d}{d u}\left(\frac{d y}{d u}\right) \frac{d u}{d x}\right] \frac{d u}{d x}+\frac{d y}{d u} \frac{d^{2} u}{d x^{2}}\\
&=\frac{d^{2} y}{d u^{2}}\left(\frac{d u}{d x}\right)^{2}+\frac{d y}{d u} \frac{d^{2} u}{d x^{2}}
\end{aligned}$$
