Step 1: Look at the structure.
We must find $\dfrac{d^2y}{dx^2}$ where $y$ is a sum of two inverse-tangent terms built from $\log x$.
Step 2: Substitute to simplify.
Let $u = \log x^2 = 2\log x$. The first term is $\tan^{-1}\left[\dfrac{1-u}{1+u}\right]$ since $\log(e/x^2)=1-u$ and $\log(ex^2)=1+u$.
Step 3: Apply a standard identity.
Using $\tan^{-1}\left(\dfrac{1-u}{1+u}\right) = \dfrac{\pi}{4} - \tan^{-1}u$, the first term becomes $\dfrac{\pi}{4} - \tan^{-1}(2\log x)$.
Step 4: Simplify the second term.
Write the second term as $\tan^{-1}\left(\dfrac{3 + 2\log x}{1 - 3\cdot 2\log x}\right) = \tan^{-1}3 + \tan^{-1}(2\log x)$.
Step 5: Add the two parts.
\[ y = \frac{\pi}{4} - \tan^{-1}(2\log x) + \tan^{-1}3 + \tan^{-1}(2\log x) = \frac{\pi}{4} + \tan^{-1}3 \] The variable piece cancels, so $y$ is a constant.
Step 6: Differentiate twice.
A constant has zero slope, so $\dfrac{dy}{dx}=0$ and therefore $\dfrac{d^2y}{dx^2}=0$. The clever move here was spotting the identity early so no messy chain-rule work was needed.
\[ \boxed{\dfrac{d^2y}{dx^2} = 0} \]