$\text { If } y=\tan ^{-1}\left(\dfrac{3-2 x}{1+6 x}\right) \text { then } \dfrac{d y}{d x} \text { is }$

The given expression is $y = \tan^{-1}\left(\dfrac{3-2x}{1+6x}\right)$.

We can rewrite the argument of the inverse tangent function as follows:

$\dfrac{3-2x}{1+6x} = \dfrac{3-2x}{1+(3)(2x)}$

Using the identity $\tan^{-1}\left(\dfrac{a-b}{1+ab}\right) = \tan^{-1}(a) - \tan^{-1}(b)$, we set $a=3$ and $b=2x$.

Thus, $y = \tan^{-1}(3) - \tan^{-1}(2x)$.

Differentiating both sides with respect to $x$:

$\dfrac{dy}{dx} = \dfrac{d}{dx}(\tan^{-1}(3)) - \dfrac{d}{dx}(\tan^{-1}(2x))$

Since $\tan^{-1}(3)$ is a constant, its derivative is $0$.

Using the chain rule for $\tan^{-1}(2x)$, we get:

$\dfrac{dy}{dx} = 0 - \dfrac{1}{1+(2x)^2} \times \dfrac{d}{dx}(2x)$

$\dfrac{dy}{dx} = -\dfrac{1}{1+4x^2} \times 2$

$\dfrac{dy}{dx} = -\dfrac{2}{1+4x^2}$

Answer: $-\dfrac{2}{1+4 x^2}$