Answer
The tangent vector at $t=1$ is ${\rm{r}}'\left( 1 \right) = \left( { - 2,1} \right)$
Please see the figure attached. The red arrow represents the tangent vector.
Work Step by Step
Recall that the derivative vector ${\rm{r}}'\left( {{t_0}} \right)$ points in the direction tangent to the path traced by ${\bf{r}}\left( t \right)$ at $t = {t_0}$. For the curve ${\bf{r}}\left( t \right) = \left( {1 - {t^2},t} \right)$ we get ${\rm{r}}'\left( t \right) = \left( { - 2t,1} \right)$. Thus, the tangent vector at $t=1$ is ${\rm{r}}'\left( 1 \right) = \left( { - 2,1} \right)$. The point corresponding to $t=1$ is ${\bf{r}}\left( 1 \right) = \left( {0,1} \right)$.
We evaluate several points for the interval $ - 3 \le t \le 3$ and list them in the following table. Then we plot the points and join them to obtain the curve.
$\begin{array}{*{20}{c}}
t&{\left( {x,y} \right)}\\
{ - 3}&{\left( { - 8, - 3} \right)}\\
{ - 2}&{\left( { - 3, - 2} \right)}\\
{ - 1}&{\left( {0, - 1} \right)}\\
0&{\left( {1,0} \right)}\\
1&{\left( {0,1} \right)}\\
2&{\left( { - 3,2} \right)}\\
3&{\left( { - 8,3} \right)}
\end{array}$
