If $y = \tan^{-1}\left(\dfrac{\log\left(\dfrac{e}{x^3}\right)}{\log ex^3}\right) + \tan^{-1}\left(\dfrac{\log(e^4x^3)}{\log\left(\dfrac{e}{x^{12}}\right)}\right)$, $x \in \left(e^{-\frac{1}{3}}, e^{\frac{1}{12}}\right)$ then $\dfrac{dy}{dx}$ is equal to...

Given function is:

$y = \tan^{-1}\left(\dfrac{\log\left(\dfrac{e}{x^3}\right)}{\log(ex^3)}\right) + \tan^{-1}\left(\dfrac{\log(e^4x^3)}{\log\left(\dfrac{e}{x^{12}}\right)}\right)$

Using the properties of logarithms, we can simplify the arguments:

$\dfrac{\log\left(\dfrac{e}{x^3}\right)}{\log(ex^3)} = \dfrac{\log e - \log x^3}{\log e + \log x^3} = \dfrac{1 - 3\log x}{1 + 3\log x}$

$\dfrac{\log(e^4x^3)}{\log\left(\dfrac{e}{x^{12}}\right)} = \dfrac{\log e^4 + \log x^3}{\log e - \log x^{12}} = \dfrac{4 + 3\log x}{1 - 12\log x}$

Substituting these back into the expression for $y$:

$y = \tan^{-1}\left(\dfrac{1 - 3\log x}{1 + 1 \cdot 3\log x}\right) + \tan^{-1}\left(\dfrac{4 + 3\log x}{1 - 4 \cdot 3\log x}\right)$

We use the inverse trigonometric identities:
$\tan^{-1}\left(\dfrac{a - b}{1 + ab}\right) = \tan^{-1}a - \tan^{-1}b$ (valid for $ab > -1$)
$\tan^{-1}\left(\dfrac{a + b}{1 - ab}\right) = \tan^{-1}a + \tan^{-1}b$ (valid for $ab < 1$)

For the first term, $a = 1$ and $b = 3\log x$. Since $x > e^{-1/3}$, we have $3\log x > -1$, so $ab > -1$.
$\tan^{-1}\left(\dfrac{1 - 3\log x}{1 + 3\log x}\right) = \tan^{-1}(1) - \tan^{-1}(3\log x) = \dfrac{\pi}{4} - \tan^{-1}(3\log x)$

For the second term, $a = 4$ and $b = 3\log x$. Since $x < e^{1/12}$, we have $12\log x < 1$, so $ab < 1$.
$\tan^{-1}\left(\dfrac{4 + 3\log x}{1 - 12\log x}\right) = \tan^{-1}(4) + \tan^{-1}(3\log x)$

Adding the two terms together:

$y = \left(\dfrac{\pi}{4} - \tan^{-1}(3\log x)\right) + \left(\tan^{-1}(4) + \tan^{-1}(3\log x)\right)$

$y = \dfrac{\pi}{4} + \tan^{-1}(4)$

Since $y$ is a constant and independent of $x$, its derivative with respect to $x$ is zero:

$\dfrac{dy}{dx} = 0$

Answer: $0$