Answer
$$\frac{3}{4\sqrt{5}}-\frac{\sqrt{5}}{4}$$
Work Step by Step
Given
$$f(x)=\int_{0}^{\sin x} \sqrt{1+t^{2}} d t \text { and } g(y)=\int_{3}^{y} f(x) d x$$
By using the Fundamental Theorem of Calculus (1)
\begin{aligned}
f'(x)&=\frac{d}{dx}\int_{0}^{\sin x} \sqrt{1+t^{2}} d t\\
&= \sqrt{1+\sin^{2}x} \frac{d}{dx}(\sin (x))\\
&= \cos (x)\sqrt{1+\sin^{2}x}
\end{aligned}
and
\begin{aligned}
g'(y)&= \frac{d}{dx}\int_3^yf(y)dy\\
&= f'(y)
\end{aligned}
Then
\begin{aligned}
g''(y)&= f''(y) \\
\text{( i.e.)}, g''(x)&= f''(x)
\end{aligned}
Hence
\begin{aligned}
g''(x)&= f''(x)\\
&= \cos(x) \frac{d}{dx}\sqrt{1+\sin^2x}+ \sqrt{1+\sin^2x}\frac{d}{dx}(\cos x)\\
&= \cos(x)\cdot \frac{1}{2\sqrt{1+\sin^2x}}\frac{d}{dx}(1+\sin^2(x))+ \sqrt{1+\sin^2x}\frac{d}{dx}(\cos(x))\\
&= \frac{2\sin \left(x\right)\cos(x)\cos \left(x\right)}{2\sqrt{1+\sin ^2\left(x\right)}}-\sin \left(x\right)\sqrt{1+\sin ^2\left(x\right)}\\
&= \frac{\sin \left(2x\right)\cos \left(x\right)}{2\sqrt{1+\sin ^2\left(x\right)}}-\sin \left(x\right)\sqrt{1+\sin ^2\left(x\right)}
\end{aligned}
It follows that
\begin{aligned}
g''(\pi /6) &= \frac{\sin \left(\pi /3\right)\cos \left(\pi /6\right)}{2\sqrt{1+\sin ^2\left(\pi /6\right)}}-\sin \left(\pi /6\right)\sqrt{1+\sin ^2\left(\pi /6\right)}\\
&= \frac{\left(\sqrt{3} /2\right) \left(\sqrt{3}/2\right)}{2\sqrt{1+ \left(1/4\right)}}-\left(1/2\right)\sqrt{1+\left(1 /4\right)}\\
&=\frac{3}{4\sqrt{5}}-\frac{\sqrt{5}}{4}
\end{aligned}
