Let $$y = y(x)$$ be the solution of the differential equation $$\dfrac{dy}{dx} + 3(\tan^2 x) \, y + 3y = \sec^2 x$$, $$y(0) = \dfrac{1}{3} + e^3$$. Then $$y\left(\dfrac{\pi}{4}\right)$$ is equal to

The differential equation is $$\dfrac{dy}{dx} + 3\tan^2 x \cdot y + 3y = \sec^2 x$$.

Rewriting: $$\dfrac{dy}{dx} + 3(\tan^2 x + 1)y = \sec^2 x$$, since $$\tan^2 x + 1 = \sec^2 x$$:

$$\dfrac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x$$

This is a linear ODE. The integrating factor is $$e^{\int 3\sec^2 x \, dx} = e^{3\tan x}$$.

Multiplying both sides: $$\dfrac{d}{dx}\left(y \cdot e^{3\tan x}\right) = \sec^2 x \cdot e^{3\tan x}$$

Integrating: $$y \cdot e^{3\tan x} = \displaystyle\int \sec^2 x \cdot e^{3\tan x} \, dx$$

Let $$t = 3\tan x$$, so $$dt = 3\sec^2 x \, dx$$:

$$y \cdot e^{3\tan x} = \displaystyle\int \dfrac{e^t}{3} \, dt = \dfrac{e^t}{3} + C = \dfrac{e^{3\tan x}}{3} + C$$

So $$y = \dfrac{1}{3} + Ce^{-3\tan x}$$.

Using $$y(0) = \dfrac{1}{3} + e^3$$: $$\dfrac{1}{3} + e^3 = \dfrac{1}{3} + C \cdot e^0$$, so $$C = e^3$$.

Therefore $$y = \dfrac{1}{3} + e^3 \cdot e^{-3\tan x}$$.

At $$x = \dfrac{\pi}{4}$$: $$\tan\dfrac{\pi}{4} = 1$$, so $$y\left(\dfrac{\pi}{4}\right) = \dfrac{1}{3} + e^3 \cdot e^{-3} = \dfrac{1}{3} + 1 = \dfrac{4}{3}$$.

Hence, the correct answer is Option B.