If $$\frac{dy}{dx} + \frac{3}{\cos^2 x}y = \frac{1}{\cos^2 x}$$, $$x \in \left(-\frac{\pi}{3}, \frac{\pi}{3}\right)$$, and $$y\left(\frac{\pi}{4}\right) = \frac{4}{3}$$, then $$y\left(-\frac{\pi}{4}\right)$$ equals:

We are asked to solve the linear first-order differential equation

$$\frac{dy}{dx}+\frac{3}{\cos ^2 x}\,y=\frac{1}{\cos ^2 x},\qquad -\frac{\pi}{3}\lt x\lt \frac{\pi}{3},$$

subject to the initial condition

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

First, we recognise the standard linear form $$\dfrac{dy}{dx}+P(x)\,y=Q(x)$$ with

$$P(x)=\frac{3}{\cos ^2 x}=3\sec ^2 x,\qquad Q(x)=\frac{1}{\cos ^2 x}=\sec ^2 x.$$

The theory of linear differential equations tells us to multiply through by an integrating factor $$\mu(x)=e^{\int P(x)\,dx}.$$

We therefore compute the exponent:

$$\int P(x)\,dx=\int 3\sec ^2 x\,dx=3\int \sec ^2 x\,dx.$$

Recalling that $$\int \sec ^2 x\,dx=\tan x,$$ we obtain

$$\int P(x)\,dx=3\tan x.$$

So the integrating factor is

$$\mu(x)=e^{3\tan x}.$$

Multiplying every term in the differential equation by this integrating factor, we have

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

By construction, the left-hand side is the derivative of the product $$\mu(x)y$$. Indeed, the product rule gives

$$\frac{d}{dx}\!\left(y\,e^{3\tan x}\right)=e^{3\tan x}\frac{dy}{dx}+3\sec ^2 x\,e^{3\tan x}\,y,$$

and since $$3\sec ^2 x=\dfrac{3}{\cos ^2 x}$$, the two expressions match perfectly. Hence we can rewrite the entire equation compactly as

$$\frac{d}{dx}\!\left(y\,e^{3\tan x}\right)=e^{3\tan x}\frac{1}{\cos ^2 x}.$$

Now we integrate both sides with respect to $$x$$:

$$\int \frac{d}{dx}\!\left(y\,e^{3\tan x}\right)\,dx=\int e^{3\tan x}\frac{1}{\cos ^2 x}\,dx.$$

The left integral is immediate, giving $$y\,e^{3\tan x}$$. For the right integral we perform the substitution

$$u=\tan x\quad\Longrightarrow\quad du=\sec ^2 x\,dx=\frac{1}{\cos ^2 x}\,dx.$$

Thus

$$\int e^{3\tan x}\frac{1}{\cos ^2 x}\,dx=\int e^{3u}\,du.$$

Integrating the exponential, we find

$$\int e^{3u}\,du=\frac{1}{3}e^{3u}+C=\frac{1}{3}e^{3\tan x}+C.$$

Putting the pieces together, we have obtained

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

where $$C$$ is the constant of integration. We now divide by $$e^{3\tan x}$$ to isolate $$y$$:

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

To determine $$C$$ we use the given initial value $$y\!\left(\dfrac{\pi}{4}\right)=\dfrac{4}{3}$$. Since $$\tan\!\left(\dfrac{\pi}{4}\right)=1$$, we substitute $$x=\dfrac{\pi}{4}$$ into the general solution:

$$\frac{4}{3}=\frac{1}{3}+C\,e^{-3\cdot1}=\frac{1}{3}+C\,e^{-3}.$$

Subtracting $$\dfrac{1}{3}$$ from both sides gives

$$\frac{4}{3}-\frac{1}{3}=C\,e^{-3}\quad\Longrightarrow\quad 1=C\,e^{-3}.$$

Hence

$$C=e^{3}.$$

Substituting this value back, the particular solution of the differential equation is

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

Finally, we evaluate this expression at $$x=-\dfrac{\pi}{4}$$. We note that $$\tan\!\left(-\dfrac{\pi}{4}\right)=-1$$. Therefore,

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

Hence, the correct answer is Option C.