$\because\quad x=e^\theta\left(\theta+\frac{1}{\theta}\right)$ and $y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

$\therefore \quad \frac{d x}{d \theta}=\frac{d}{d \theta}\left[e^\theta \cdot\left(\theta+\frac{1}{\theta}\right)\right]$

$$\begin{aligned} & =e^\theta \cdot \frac{d}{d \theta}\left(\theta+\frac{1}{\theta}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot \frac{d}{d \theta} e^\theta \\ & =e^\theta\left(1-\frac{1}{\theta^2}\right)+\left(\theta+\frac{1}{\theta}\right) e^\theta \\ & =e^\theta\left(1-\frac{1}{\theta^2}+\theta+\frac{1}{\theta}\right) \\ & =e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)\text{.... (i)} \end{aligned}$$

and $$\frac{d y}{d \theta}=\frac{d}{d \theta}\left[e^{-\theta} \cdot\left(\theta-\frac{1}{\theta}\right)\right]$$

$$\begin{aligned} & =e^{-\theta} \cdot \frac{d}{d \theta}\left(\theta-\frac{1}{\theta}\right)+\frac{d}{d \theta} e^{-\theta}\left(\theta-\frac{1}{\theta}\right) \\ & =e^{-\theta}\left(1+\frac{1}{\theta^2}\right)+\left(\theta-\frac{1}{\theta}\right) e^{-\theta} \cdot \frac{d}{d \theta}(-\theta) \\ & =e^{-\theta}\left[\frac{\theta^2+1}{\theta^2}-\frac{\theta^2-1}{\theta}\right]=e^{-\theta}\left[\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right]\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}\left(\frac{\theta^2+1-\theta^3+\theta}{\theta^2}\right)}{e^\theta\left(\frac{\theta^2-1+\theta^3+\theta}{\theta^2}\right)} \\ & =e^{-2 \theta}\left(\frac{-\theta^3+\theta^2+\theta+1}{\theta^3+\theta^2+\theta-1}\right) \end{aligned}$$

$$\begin{aligned} \because\quad x & =3 \cos \theta-2 \cos ^3 \theta \text { and } y=3 \sin \theta-2 \sin ^3 \theta \\ \therefore\quad \frac{d x}{d \theta} & =\frac{d}{d \theta}(3 \cos \theta)-\frac{d}{d \theta}\left(2 \cos ^3 \theta\right) \\ & =3 \cdot(-\sin \theta)-2 \cdot 3 \cos ^2 \theta \cdot \frac{d}{d \theta} \cdot \cos \theta \\ & =-3 \sin \theta+6 \cos ^2 \theta \sin \theta \\ \text{and}\quad\frac{d y}{d \theta} & =3 \cos A-2 \cdot 3 \sin ^2 \theta \cdot \frac{d}{d \theta} \cdot \sin \theta \\ & =3 \cos \theta-6 \sin ^2 \theta \cdot \cos \theta \\ \text{Now,}\quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{3 \cos \theta-6 \sin ^2 \theta \cos \theta}{-3 \sin \theta+6 \cos ^2 \theta \sin \theta} \\ & =\frac{3 \cos \theta\left(1-2 \sin ^2 \theta\right)}{3 \sin \theta\left(-1+2 \cos ^2 \theta\right)}=\cot \theta \cdot \frac{\cos 2 \theta}{\cos 2 \theta}=\cot \theta \end{aligned}$$

$$\begin{array}{ll} \because & \sin x=\frac{2 t}{1+t^2} \quad\text{.... (i)}\\ \text { and } & \tan y=\frac{2 t}{1-t^2}\quad\text{.... (ii)} \end{array}$$

$$\begin{aligned} & \begin{aligned} \therefore \quad \frac{d}{d x} \sin x \cdot \frac{d x}{d t} & =\frac{d}{d t}\left(\frac{2 t}{1+t^2}\right) \\ \Rightarrow \quad \cos x \frac{d x}{d t} & =\frac{\left(1+t^2\right) \cdot \frac{d}{d t}(2 t)-(2 t) \cdot \frac{d}{d t}\left(1+t^2\right)}{\left(1+t^2\right)^2} \\ & =\frac{2\left(1+t^2\right)-2 t \cdot 2 t}{\left(1+t^2\right)^2}=\frac{2+2 t^2-4 t^2}{\left(1+t^2\right)^2} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\cos x} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\sqrt{1-\sin ^2 x}}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{1}{\sqrt{1-\left(\frac{2 t}{\left.1+t^2\right)^2}\right.}} \end{aligned} \end{aligned}$$

$\Rightarrow \quad \frac{d x}{d t}=\frac{2\left(1-t^2\right)}{\left(1+t^2\right)^2} \cdot \frac{\left(1+t^2\right)}{\left(1-t^2\right)}=\frac{2}{1+t^2}\quad\text{..... (iii)}$

$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} \frac{d}{d y} \tan y \cdot \frac{d y}{d t} & =\frac{d}{d t}\left(\frac{2 t}{1-t^2}\right) \\ \sec ^2 y \frac{d y}{d t} & =\frac{\left(1-t^2\right) \frac{d}{d t} \cdot(2 t)-2 t \cdot \frac{d}{d t}\left(1-t^2\right)}{\left(1-t^2\right)^2} \\ \frac{d y}{d t} & =\frac{2-2 t^2+4 t^2}{\left(1-t^2\right)^2} \cdot \frac{1}{\sec ^2 y} \\ & =\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{\left(1+\tan ^2 y\right)}=\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{1}{1+\frac{4 t^2}{\left(1-t^2\right)^2}} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{2\left(1+t^2\right)}{\left(1-t^2\right)^2} \cdot \frac{\left(1-t^2\right)^2}{\left(1+t^2\right)^2}=\frac{2}{1+t^2} \quad\text{.... (iv)}\\ \therefore \quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{2 / 1+t^2}{2 / 1+t^2}=1\quad\text{[from Eqs. (iii) and (iv)]} \end{aligned}$$

$$\begin{aligned} & \because \quad x=\frac{1+\log t}{t^2} \text { and } y=\frac{3+2 \log t}{t} \\ & \therefore \quad \frac{d x}{d t}=\frac{t^2 \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t} t^2}{\left(t^2\right)^2} \end{aligned}$$

$$\begin{aligned} & =\frac{t^2 \cdot \frac{1}{t}-(1+\log t) \cdot 2 t}{t^4}=\frac{t-(1+\log t) \cdot 2 t}{t^4} \\ & =\frac{t}{t^4}\left[1-2(1+\log t)=\frac{-1-2 \log t}{t^3}\right. \quad\text{.... (i)}\\ \text{and}\quad \frac{d y}{d t} & =\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t} t}{t^2} \\ & =\frac{t \cdot 2 \cdot \frac{1}{t}-(3+2 \log t) \cdot 1}{t^2} \\ & =\frac{2-3-2 \log t}{t^2}=\frac{-1-2 \log t}{t^2} \quad \quad\text{.... (ii)}\\ \therefore\quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{-1-2 \log t / t^2}{-1-2 \log t / t^3}=t \end{aligned}$$

$$\begin{aligned} & \because \quad x=e^{\cos 2 t} \text { and } y=e^{\sin 2 t} \\ & \therefore \quad \frac{d x}{d t}=\frac{d}{d t} e^{\cos 2 t}=e^{\cos 2 t} \cdot \frac{d}{d t} \cos 2 t \\ & =e^{\cos 2 t} \cdot(-\sin 2 t) \cdot \frac{d}{d t}(2 t) \end{aligned}$$

$$\begin{aligned} \frac{d x}{d t} & =-2 e^{\cos 2 t} \cdot \sin 2 t \quad\text{.... (i)}\\ \text{and}\quad \frac{d y}{d t} & =\frac{d}{d t} e^{\sin 2 t}=e^{\sin 2 t} \cdot \frac{d}{d t} \sin 2 t \\ & =e^{\sin 2 t} \cos 2 t \cdot \frac{d}{d t} 2 t \\ & =2 e^{\sin 2 t} \cdot \cos 2 t\quad\text{.... (ii)} \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{2 e^{\sin 2 t} \cdot \cos 2 t}{-2 e^{\cos 2 t} \cdot \sin 2 t} \\ & =\frac{e^{\sin 2 t} \cdot \cos 2 t}{e^{\cos 2 t} \cdot \sin 2 t}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We know that, }\quad \log x & =\cos 2 t \cdot \log e=\cos 2 t \quad\text{.... (iv)}\\ \text { and }\quad \log y & =\sin 2 t \cdot \log e=\sin 2 t \quad\text{..... (v)}\\ \therefore\quad \frac{d y}{d x} & =\frac{-y \log x}{x \log y}\quad \text { [using Eqs. (iv) and (v) in Eq. (iii) and } x=e^{\cos 2 t}, y=e^{\sin 2 t} \text { ] } \end{aligned} \end{aligned}$$

Hence proved.