Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t=0$:
${\bf{N}}\left( 0 \right) = \left( {\frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\cosh t,\sinh t,{t^2}} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {\sinh t,\cosh t,2t} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {\cosh t,\sinh t,2} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 4{t^2}} $
$v'\left( t \right) = \frac{{2\sinh t\cosh t + 2\cosh t\sinh t + 8t}}{{2\sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 4{t^2}} }}$
$v'\left( t \right) = \frac{{2\sinh t\cosh t + 4t}}{{\sqrt {{{\sinh }^2}t + {{\cosh }^2}t + 4{t^2}} }}$
Evaluate $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t=0$:
$v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right) = \left( {1,0,2} \right) - 0\left( {0,1,0} \right)$
$v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right) = \left( {1,0,2} \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t=0$ is given by
${\bf{N}}\left( 0 \right) = \frac{{v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right)}}{{||v\left( 0 \right){\bf{r}}{\rm{''}}\left( 0 \right) - v'\left( 0 \right){\bf{r}}'\left( 0 \right)||}}$
${\bf{N}}\left( 0 \right) = \frac{{\left( {1,0,2} \right)}}{{||\left( {1,0,2} \right)||}}$
${\bf{N}}\left( 0 \right) = \left( {\frac{1}{{\sqrt 5 }},0,\frac{2}{{\sqrt 5 }}} \right)$
