Answer: (i) \(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\cot y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\).

(ii) \(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{4}{3\sqrt{3}}=-\dfrac{4\sqrt{3}}{9}\).

(i) From \(\cos y=x\), differentiate implicitly with respect to \(x\):

\(-\sin y\,\dfrac{\mathrm dy}{\mathrm dx}=1\), so \(\dfrac{\mathrm dy}{\mathrm dx}=-(\sin y)^{-1}\).

Differentiate again:

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=(\sin y)^{-2}\cos y\,\dfrac{\mathrm dy}{\mathrm dx}\).

Since \(\dfrac{\mathrm dy}{\mathrm dx}=-(\sin y)^{-1}\), this becomes

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{\cos y}{\sin^3y}=-\cot y\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2\),

as required.

(ii) At \(y=\dfrac{\pi}{3}\),

\(\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{1}{\sin(\pi/3)}=-\dfrac{2}{\sqrt3}\), and \(\cot(\pi/3)=\dfrac{1}{\sqrt3}\).

Therefore

\(\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{1}{\sqrt3}\left(-\dfrac{2}{\sqrt3}\right)^2=-\dfrac{4}{3\sqrt3}=-\dfrac{4\sqrt3}{9}\).