(i) Given that
[tex]\tan^{-1}(x) + \tan^{-1}(y) + \tan^{-1}(xy) = \dfrac{7\pi}{12}[/tex]
when [tex]x=1[/tex] this reduces to
[tex]\tan^{-1}(1) + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]
[tex]\dfrac\pi4 + 2 \tan^{-1}(y) = \dfrac{7\pi}{12}[/tex]
[tex]2 \tan^{-1}(y) = \dfrac\pi3[/tex]
[tex]\tan^{-1}(y) = \dfrac\pi6[/tex]
[tex]\tan\left(\tan^{-1}(y)\right) = \tan\left(\dfrac\pi6\right)[/tex]
[tex]\implies \boxed{y = \dfrac1{\sqrt3}}[/tex]
(ii) Differentiate [tex]\tan^{-1}(xy)[/tex] implicitly with respect to [tex]x[/tex]. By the chain and product rules,
[tex]\dfrac d{dx} \tan^{-1}(xy) = \dfrac1{1+(xy)^2} \times \dfrac d{dx}xy = \boxed{\dfrac{y + x\frac{dy}{dx}}{1 + x^2y^2}}[/tex]
(iii) Differentiating both sides of the given equation leads to
[tex]\dfrac1{1+x^2} + \dfrac1{1+y^2} \dfrac{dy}{dx} + \dfrac{y + x\frac{dy}{dx}}{1+x^2y^2} = 0[/tex]
where we use the result from (ii) for the derivative of [tex]\tan^{-1}(xy)[/tex].
Solve for [tex]\frac{dy}{dx}[/tex] :
[tex]\dfrac1{1+x^2} + \left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} + \dfrac y{1+x^2y^2} = 0[/tex]
[tex]\left(\dfrac1{1+y^2} + \dfrac x{1+x^2y^2}\right) \dfrac{dy}{dx} = -\left(\dfrac1{1+x^2} + \dfrac y{1+x^2y^2}\right)[/tex]
[tex]\dfrac{1+x^2y^2 + x(1+y^2)}{(1+y^2)(1+x^2y^2)} \dfrac{dy}{dx} = - \dfrac{1+x^2y^2 + y(1+x^2)}{(1+x^2)(1+x^2y^2)}[/tex]
[tex]\implies \dfrac{dy}{dx} = - \dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2) (1 + x^2y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2) (1+x^2y^2)}[/tex]
[tex]\implies \dfrac{dy}{dx} = -\dfrac{(1 + x^2y^2 + y + x^2y) (1 + y^2)}{(1 + x^2y^2 + x + xy^2) (1+x^2)}[/tex]
From part (i), we have [tex]x=1[/tex] and [tex]y=\frac1{\sqrt3}[/tex], and substituting these leads to
[tex]\dfrac{dy}{dx} = -\dfrac{\left(1 + \frac13 + \frac1{\sqrt3} + \frac1{\sqrt3}\right) \left(1 + \frac13\right)}{\left(1 + \frac13 + 1 + \frac13\right) \left(1 + 1\right)}[/tex]
[tex]\dfrac{dy}{dx} = -\dfrac{\left(\frac43 + \frac2{\sqrt3}\right) \times \frac43}{\frac83 \times 2}[/tex]
[tex]\dfrac{dy}{dx} = -\dfrac13 - \dfrac1{2\sqrt3}[/tex]
as required.