1. Given differential equation

$\frac{dy}{dx}+\frac{3}{\cos^2 x}y=\frac{1}{\cos^2 x}$

Since $\dfrac{1}{\cos^2 x}=\sec^2 x$, rewrite it as

$\frac{dy}{dx}+3\sec^2 x\,y=\sec^2 x$

This is a linear differential equation of the form

$\frac{dy}{dx}+P(x)y=Q(x)$

with

$P(x)=3\sec^2 x, \qquad Q(x)=\sec^2 x.$

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2. Find the integrating factor

$\text{I.F.}=e^{\int P(x)dx}=e^{\int 3\sec^2 x\,dx}=e^{3\tan x}$

because

$\int \sec^2 x\,dx=\tan x.$

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3. Multiply the equation by the integrating factor

$e^{3\tan x}\frac{dy}{dx}+3\sec^2 x\,e^{3\tan x}y=\sec^2 x\,e^{3\tan x}$

The left side becomes

$\frac{d}{dx}\left(ye^{3\tan x}\right)=\sec^2 x\,e^{3\tan x}.$

So,

$\frac{d}{dx}\left(ye^{3\tan x}\right)=\sec^2 x\,e^{3\tan x}.$

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4. Integrate both sides

$ye^{3\tan x}=\int \sec^2 x\,e^{3\tan x}\,dx + C$

Let

$t=3\tan x \implies dt=3\sec^2 x\,dx \implies \sec^2 x\,dx=\frac{dt}{3}.$

Hence,

$\int \sec^2 x\,e^{3\tan x}\,dx=\frac{1}{3}\int e^t dt=\frac{1}{3}e^t=\frac{1}{3}e^{3\tan x}.$

Therefore,

$ye^{3\tan x}=\frac{1}{3}e^{3\tan x}+C$

and so

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

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5. Use the initial condition

Given

$y\left(\frac{\pi}{4}\right)=\frac{4}{3}$

and

$\tan\frac{\pi}{4}=1.$

Substitute into the solution:

$\frac{4}{3}=\frac{1}{3}+Ce^{-3}$

$Ce^{-3}=1$

$C=e^3.$

So the particular solution is

$y=\frac{1}{3}+e^3e^{-3\tan x}=\frac{1}{3}+e^{3-3\tan x}.$

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6. Find $y\left(-\frac{\pi}{4}\right)$

Since

$\tan\left(-\frac{\pi}{4}\right)=-1,$

we get

$y\left(-\frac{\pi}{4}\right)=\frac{1}{3}+e^{3-3(-1)}=\frac{1}{3}+e^6.$

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7. Compare with options

Thus,

$y\left(-\frac{\pi}{4}\right)=\frac{1}{3}+e^6$

So the correct option is A.