Nothing to show here but you have a lot to “show”!

(i) $y=\sqrt{1+\ln (1+\sin 2 x)}$
$y^2=1+\ln (1+\sin 2 x)$
$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{2 \cos 2 x}{1+\sin 2 x}$
$y \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\cos 2 x}{1+\sin 2 x}$

(ii) $\begin{aligned} y \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2 &=\frac{-2 \sin 2 x(1+\sin 2 x)-2 \cos 2 x(\cos 2 x)}{(1+\sin 2 x)^2} \\ &=\frac{-2 \sin 2 x-2 \sin ^2 2 x-2 \cos ^2 2 x}{(1+\sin 2 x)^2} \\ &=\frac{-2(\sin 2 x+1)}{(1+\sin 2 x)^2}=\frac{-2}{(1+\sin 2 x)} \end{aligned}$

(iii) $y \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}+\frac{\mathrm{d} y}{\mathrm{~d} x} \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right) \frac{\mathrm{d}^2 y}{d x^2}=\frac{4 \cos 2 x}{(1+\sin 2 x)^2}$
Let $\mathrm{f}(x)=\sqrt{1+\ln (1+\sin 2 x)}$
$\mathrm{f}(0)=1$
$\mathrm{f}^{\prime}(0)=$
$\mathrm{f}^{\prime \prime}(0)=-3$
$\mathrm{f}^{\prime \prime \prime}(0)=13$
The Maclaurin expansion of
$y=\sqrt{1+\ln (1+\sin 2 x)}=1+x-\frac{3 x^2}{2 !}+\frac{13 x^3}{3 !}+\ldots$

(iv) 

$\left(1+x-\frac{3}{2} x^2+\frac{13}{6} x^3+\ldots\right)^2
=\left(1+x-\frac{3}{2} x^2+\frac{13}{6} x^3+\ldots\right)\left(1+x-\frac{3}{2} x^2+\frac{13}{6} x^3+\ldots\right)$
$=1+x-\frac{3}{2} x^2+\frac{13}{6} x^3$
$\quad+x+x^2-\frac{3}{2} x^3$
$\quad-\frac{3}{2} x^2-\frac{3}{2} x^3$
$\quad+\frac{13}{6} x^3+\ldots. $

$=1+2 x-2 x^2+\frac{4}{3} x^3+…$
Using standard series expansion from MF 26,
$y^2 =1+\ln (1+\sin 2 x)=1+\ln \left[1+\left(5^5 2 x-\frac{(2 x)^3}{3 !}+\ldots\right)\right] $

$=1+\left(2 x-\frac{8 x^3}{6}\right)\left(-\frac{1}{2}\left({ }^2-2 x-\frac{8 x^3}{6}\right)^2+\frac{1}{3}\left(2 x-\frac{8 x^3}{6}\right)^3+\ldots\right.$
$=1+2 x=\frac{4}{3} x^3-\frac{1}{2}(2 x)^2+\frac{1}{3}(2 x)^3+. . $
$=1+2 x-2 x^2+\frac{4}{3} x^3+\ldots$