Solve this

$\operatorname{as} x=a e^{\theta}(\sin \theta-\cos \theta)$

Differentiating it with respect to $\theta$

$\frac{d x}{d \theta}=a\left[e^{\theta} \frac{d(\sin \theta-\cos \theta)}{d \theta}+(\sin \theta-\cos \theta) \frac{d\left(e^{\theta}\right)}{d \theta}\right]$

$=a\left[e^{\theta}(\cos \theta+\sin \theta)+(\sin \theta-\cos \theta) e^{\theta}\right]$

$\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left[2 \mathrm{e}^{\theta} \sin \theta\right]$      .....(1)

And, $y=a e^{\theta}(\sin \theta+\cos \theta)$

Differentiating it with respect to $\theta$,

$\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}\left[\mathrm{e}^{\theta} \frac{\mathrm{d}(\sin \theta+\cos \theta)}{\mathrm{d} \theta}+(\sin \theta+\cos \theta) \frac{\mathrm{d}\left(\mathrm{e}^{\theta}\right)}{\mathrm{d} \theta}\right]$

$=a\left[e^{\theta}(\cos \theta-\sin \theta)+(\sin \theta+\cos \theta) e^{\theta}\right]$

$\frac{d y}{d \theta}=a\left[2 e^{\theta} \cos \theta\right] \ldots \ldots(2)$

Dividing equation (2) by equation (1),

$\frac{d y}{d x}=\frac{a\left(2 e^{\theta} \cos \theta\right)}{a\left(2 e^{\theta} \sin \theta\right)}$

$\frac{d y}{d x}=\cot \theta$