Given, $x=c^\theta \sin \theta, y=c^\theta \cos \theta$
We have, $\frac{d x}{d \theta}=c^0 \cos \theta+e^\theta \sin \theta$
$$
\text { and } \begin{aligned}
& =c^\theta(\cos \theta+\sin \theta) \\
d \theta & =c^\theta(-\sin \theta)+c^\theta \cos \theta \\
& =c^\theta(\cos \theta-\sin \theta)
\end{aligned}
$$
On dividing Eq. (ii) by Eq (i), we get
$$
\begin{aligned}
\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}} & =\frac{e^\theta(\cos \theta-\sin \theta)}{e^\theta(\cos \theta+\sin \theta)} \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{\cos \theta\left(1-\frac{\sin \theta}{\cos \theta}\right)}{\cos \theta\left(1+\frac{\sin \theta}{\cos \theta}\right)}=\frac{1-\tan \theta}{1+\tan \theta}
\end{aligned}
$$
As we also have, $\frac{x}{y}=\frac{e^\theta \sin \theta}{e^\theta \cos \theta} \Rightarrow \frac{x}{y}=\tan \theta$
At $\quad(x, y)=(1,1) \Rightarrow \tan \theta=\frac{1}{1}=1$
Putting $\tan \theta=1$ into Eq. (iii), we get
$$
\frac{d y}{d x}=\frac{1-1}{1+1}=\frac{0}{2}=0
$$