Answer
$ a.\quad$ see image
$ b.\quad \mathrm{r}^{\prime}(t)=(e^{t}) \mathrm{i}-(e^{-t})\mathrm{j}$,
$ c.\quad$ see image
Work Step by Step
$a.$
Parametric equations: $\mathrm{r}(t):\quad\left\{\begin{array}{ll}
x=e^{t} & \\
y=e^{-t} &
\end{array}\right.$
comparing y with x: $\displaystyle \quad y=x^{-1}=\frac{1}{x}$
and, since both x and y are positive, this is the part of the hyperbola in the 1st quadrant
see image
$b.$
Differentiate $(\displaystyle \frac{d}{dt})$ each component function
$\mathrm{r}^{\prime}(t):\quad\left\{\begin{array}{l}
x=e^{t}\\
y=-e^{-t}
\end{array}\right.$
$\mathrm{r}^{\prime}(t)=(e^{t}) \mathrm{i}-(e^{-t})\mathrm{j}$,
$c. $
For $t_{o}$ = $0,$
position vector : black$\quad \mathrm{r}(t_{o})$=$\langle 1,1\rangle$
tangent vector: red$\quad \mathrm{r}^{\prime}(t_{o})$=$\langle 1,-1\rangle$
