Solve the following

Given,

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

Differentiating both the parametric functions w.r.t. $\theta$.

$\frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$

$\frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}+\theta+\frac{1}{\theta}\right) \Rightarrow e^{\theta}\left(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}\right)$

$=\frac{e^{\theta}\left(\theta^{3}+\theta^{2}+\theta-1\right)}{\theta^{2}}$

$y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$

$\frac{d y}{d \theta}=e^{-\theta}\left(1+\frac{1}{\theta^{2}}\right)+\left(\theta-\frac{1}{\theta}\right) \cdot\left(-e^{-\theta}\right)$

$\frac{d y}{d \theta}=e^{-\theta}\left(1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}\right) \Rightarrow e^{-\theta}\left(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}\right)$

$=e^{-\theta} \frac{\left(-\theta^{3}+\theta^{2}+\theta+1\right)}{\theta^{2}}$

$\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}}\right)}{e^{\theta}\left(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}}\right)}$

$=e^{-2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)$

Thus, $\quad \frac{d y}{d x}=e^{-2 \theta}\left(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}\right)$