If $$y = y(x)$$ is the solution of the differential equation $$\frac{dy}{dx} = (\tan x - y)\sec^2 x$$, $$x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, such that $$y(0) = 0$$, then $$y\left(-\frac{\pi}{4}\right)$$ is equal to:

We have the first-order differential equation

$$\frac{dy}{dx}= (\tan x - y)\sec^{2}x, \qquad x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$$

First, we expand the right-hand side to put the equation in the standard linear form $$\dfrac{dy}{dx}+P(x)\,y=Q(x).$$

Multiplying out,

$$\frac{dy}{dx}= \tan x\sec^{2}x - y\sec^{2}x.$$

Now we bring the term containing $$y$$ to the left:

$$\frac{dy}{dx}+y\sec^{2}x=\tan x\sec^{2}x.$$

Comparing with the standard form, we identify

$$P(x)=\sec^{2}x,\qquad Q(x)=\tan x\sec^{2}x.$$

For a linear equation, the integrating factor is given by

$$\text{IF}=e^{\int P(x)\,dx}.$$

Since $$P(x)=\sec^{2}x$$ and we know $$\int\sec^{2}x\,dx=\tan x,$$ we obtain

$$\text{IF}=e^{\tan x}.$$

Multiplying the entire differential equation by this integrating factor, we get

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

The left-hand side is now the exact derivative of $$y e^{\tan x}$$ with respect to $$x$$, because

$$\frac{d}{dx}\bigl(y e^{\tan x}\bigr)=e^{\tan x}\frac{dy}{dx}+y\sec^{2}x\,e^{\tan x}.$$

Therefore, we can write

$$\frac{d}{dx}\bigl(y e^{\tan x}\bigr)=\tan x\sec^{2}x\,e^{\tan x}.$$

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

$$\int \frac{d}{dx}\bigl(y e^{\tan x}\bigr)\,dx=\int \tan x\sec^{2}x\,e^{\tan x}\,dx.$$

The left integral is simply

$$y e^{\tan x}+C_{1},$$

but we absorb the constant of integration into a single constant later. For the right integral, we make the substitution

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

This changes the integral to

$$\int u e^{u}\,du.$$

We now recall the standard formula

$$\int u e^{u}\,du=(u-1)e^{u}+C_{2}.$$

Reverting to $$u=\tan x$$, we have

$$\int \tan x\sec^{2}x\,e^{\tan x}\,dx=(\tan x-1)e^{\tan x}+C_{2}.$$

Putting both sides together and renaming the constant $$C_{2}-C_{1}$$ simply as $$C$$, we obtain

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

Dividing by $$e^{\tan x}$$ gives an explicit expression for $$y$$:

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

We now use the initial condition $$y(0)=0$$. At $$x=0$$ we have $$\tan 0=0$$ and $$e^{-\tan 0}=e^{0}=1$$, hence

$$0=\bigl(\tan 0\bigr)-1+C e^{-\tan 0}\; \Longrightarrow\; 0=0-1+C(1).$$

So

$$C=1.$$

Therefore the particular solution satisfying the given condition is

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

We are asked to find $$y\!\left(-\dfrac{\pi}{4}\right).$$ First compute $$\tan\!\left(-\dfrac{\pi}{4}\right)=-1.$$ Substituting into the expression for $$y$$, we get

$$y\!\left(-\frac{\pi}{4}\right)=\bigl(-1\bigr)-1+e^{-(-1)}=-2+e^{1}=e-2.$$

On comparing with the given options, we see that $$e-2$$ corresponds to Option C (Option 3 in the list).

Hence, the correct answer is Option C.