(a) $m=-\frac{1}{3} $

(b)$ x^2-y^2=2, x \geq \sqrt{2}$

(a)
$$
\begin{array}{l}
x=\frac{1}{2}\left(e^{3 t}+2 e^{-3 t}\right) \Rightarrow \frac{d x}{d t}=\frac{1}{2}\left(3 e^{3 t}-6 e^{-3 t}\right) \\
y=\frac{1}{2}\left(e^{3 t}-2 e^{-3 t}\right) \Rightarrow \frac{d y}{d t}=\frac{1}{2}\left(3 e^{3 t}+6 e^{-3 t}\right) \\
\frac{d y}{d x}=\frac{3 e^{3 t}+6 e^{-3 t}}{3 e^{3 t}-6 e^{-3 t}} \\
\left.\frac{d y}{d x}\right|_{t=\frac{1}{3} \ln 2}=\frac{3 e^{\ln 2}+6 e^{-\ln 2}}{3 e^{\ln 2}-6 e^{-\ln 2}}=3 \Rightarrow m_{\text {normal }}=-\frac{1}{3}
\end{array}
$$

(b)
$$
\begin{array}{l}
x=\frac{1}{2}\left(e^{3 t}+2 e^{-3 t}\right) \Rightarrow x^2=\frac{1}{4}\left(e^{6 t}+4+4 e^{-6 t}\right) \\
y=\frac{1}{2}\left(e^{3 t}-2 e^{-3 t}\right) \Rightarrow y^2=\frac{1}{4}\left(e^{6 t}-4+4 e^{-6 t}\right) \\
x^2-y^2=2, x \geq \sqrt{2}
\end{array}
$$