Given parametric equations

$x=3\tan t, \qquad y=3\sec t$

We need to find:

$\frac{d^2y}{dx^2}$

at $t=\frac{\pi}{4}$.

First derivative for parametric form

For parametric equations,

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

Compute derivatives:

$\frac{dx}{dt}=3\sec^2 t$

$\frac{dy}{dt}=3\sec t\tan t$

Therefore,

$\frac{dy}{dx}=\frac{3\sec t\tan t}{3\sec^2 t}=\frac{\tan t}{\sec t}=\sin t$

Second derivative for parametric form

Now,

$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}(\sin t)}{\frac{dx}{dt}}$

Since

$\frac{d}{dt}(\sin t)=\cos t$

and

$\frac{dx}{dt}=3\sec^2 t$

we get

$\frac{d^2y}{dx^2}=\frac{\cos t}{3\sec^2 t}$

Using $\sec^2 t=\frac{1}{\cos^2 t}$,

$\frac{d^2y}{dx^2}=\frac{\cos t}{3\cdot \frac{1}{\cos^2 t}}=\frac{\cos^3 t}{3}$

Evaluate at $t=\frac{\pi}{4}$

$\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}$

So,

$\frac{d^2y}{dx^2}=\frac{1}{3}\left(\frac{1}{\sqrt{2}}\right)^3$

$=\frac{1}{3}\cdot \frac{1}{2\sqrt{2}}$

$=\frac{1}{6\sqrt{2}}$

Match with options

$\boxed{\frac{1}{6\sqrt{2}}}$

This is Option B.