$\begin{aligned} & x=e^t(\sin t-\cos t) \text { and } y=e^t(\sin t+\cos t) \\ & \therefore \quad \frac{d x}{d t}=e^t(\sin t-\cos t)+e^t(\cos t+\sin t)=2 e^t \sin t \\ & \frac{d y}{d t}=e^t(\sin t+\cos t)+e^t(\cos t-\sin t)=2 e^t \sin t \\ & \therefore \quad \frac{d y}{d x}=\frac{2 e^t \cos t}{2 e^t \sin t} \Rightarrow\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{3}}=\cot \left(\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}} \\ & \end{aligned}$