The user wants me to solve a first-order linear differential equation and verify four given statements about the solution and its derivative.

Step 1: Identify the type of differential equation and find the integrating factor. The given differential equation is: $y' - y\tan x = 2x\sec x$ This is a linear first-order differential equation of the form $y' + P(x)y = Q(x)$, where $P(x) = -\tan x$ and $Q(x) = 2x\sec x$.

The integrating factor (I.F.) is given by $e^{\int P(x) dx}$. $I.F. = e^{\int -\tan x \,dx} = e^{-(-\ln|\cos x|)} = e^{\ln|\cos x|} = |\cos x|$ Since the initial condition is given at $x=0$, we consider an interval around $x=0$, such as $(-\pi/2, \pi/2)$, where $\cos x > 0$. So, we can take the integrating factor as $I.F. = \cos x$.

Step 2: Solve the differential equation. The solution of the linear differential equation is given by $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) \,dx + C$. $y \cos x = \int (2x \sec x) (\cos x) \,dx + C$ $y \cos x = \int 2x \,dx + C$ $y \cos x = 2 \frac{x^2}{2} + C$ $y \cos x = x^2 + C$

Step 3: Apply the initial condition to find the constant C. We are given the initial condition $y(0) = 0$. Substituting $x=0$ and $y=0$ into the general solution: $0 \cdot \cos(0) = 0^2 + C$ $0 \cdot 1 = 0 + C$ $C = 0$

Step 4: Determine the particular solution and its derivative. Substituting $C=0$ back into the general solution, we get the particular solution: $y \cos x = x^2$ $y(x) = \frac{x^2}{\cos x} = x^2 \sec x$

Now, we find the derivative, $y'(x)$. We can use the original differential equation: $y' = y\tan x + 2x\sec x$. Substituting the expression for $y(x)$: $y'(x) = (x^2 \sec x) \tan x + 2x \sec x$ $y'(x) = 2x \sec x + x^2 \sec x \tan x$

Step 5: Evaluate each option.

Option A: $y\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {8\sqrt 2 }}$ $y\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^2 \sec\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16} \cdot \sqrt{2} = \frac{\sqrt{2}\pi^2}{16}$ To match the option's format, we can write $\frac{\sqrt{2}\pi^2}{16} = \frac{\pi^2}{16/\sqrt{2}} = \frac{\pi^2}{16\sqrt{2}/2} = \frac{\pi^2}{8\sqrt{2}}$. Thus, option A is correct.

Option B: $y'\left( {{\pi \over 4}} \right) = {{{\pi ^2}} \over {18}}$ $y'\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2 \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right)$ $y'\left(\frac{\pi}{4}\right) = \frac{\pi}{2}(\sqrt{2}) + \frac{\pi^2}{16}(\sqrt{2})(1) = \frac{\sqrt{2}\pi}{2} + \frac{\sqrt{2}\pi^2}{16}$ This is clearly not equal to $\frac{\pi^2}{18}$. Thus, option B is incorrect.

Option C: $y\left( {{\pi \over 3}} \right) = {{{\pi ^2}} \over 9}$ $y\left(\frac{\pi}{3}\right) = \left(\frac{\pi}{3}\right)^2 \sec\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9} \cdot 2 = \frac{2\pi^2}{9}$ This is not equal to $\frac{\pi^2}{9}$. Thus, option C is incorrect.

Option D: $y'\left( {{\pi \over 3}} \right) = {{4\pi } \over 3} + {{2{\pi ^2}} \over {3\sqrt 3 }}$ $y'\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) + \left(\frac{\pi}{3}\right)^2 \sec\left(\frac{\pi}{3}\right) \tan\left(\frac{\pi}{3}\right)$ $y'\left(\frac{\pi}{3}\right) = \frac{2\pi}{3}(2) + \frac{\pi^2}{9}(2)(\sqrt{3}) = \frac{4\pi}{3} + \frac{2\sqrt{3}\pi^2}{9}$ Let's check if this matches the expression in the option. The second term in the option is $\frac{2\pi^2}{3\sqrt{3}}$. Rationalizing the denominator: $\frac{2\pi^2}{3\sqrt{3}} = \frac{2\pi^2 \cdot \sqrt{3}}{3\sqrt{3} \cdot \sqrt{3}} = \frac{2\sqrt{3}\pi^2}{9}$. This matches our calculated result. Thus, option D is correct.

Conclusion: The correct options are A and D.