Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t=1$ is
${\bf{N}}\left( 1 \right) = \left( {0,\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {\frac{{{t^2}}}{2},\frac{{{t^3}}}{3},t} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {t,{t^2},1} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {1,2t,0} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {{t^4} + {t^2} + 1} $
$v'\left( t \right) = \frac{{4{t^3} + 2t}}{{2\sqrt {{t^4} + {t^2} + 1} }} = \frac{{2{t^3} + t}}{{\sqrt {{t^4} + {t^2} + 1} }}$
Evaluate $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t=1$:
$v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right) = \sqrt 3 \left( {1,2,0} \right) - \frac{3}{{\sqrt 3 }}\left( {1,1,1} \right)$
$v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right) = \left( {0,\sqrt 3 , - \sqrt 3 } \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t=1$ is given by
${\bf{N}}\left( 1 \right) = \frac{{v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right)}}{{||v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right)||}}$
${\bf{N}}\left( 1 \right) = \frac{{\left( {0,\sqrt 3 , - \sqrt 3 } \right)}}{{||\left( {0,\sqrt 3 , - \sqrt 3 } \right)||}} = \frac{1}{{\sqrt 6 }}\left( {0,\sqrt 3 , - \sqrt 3 } \right)$
${\bf{N}}\left( 1 \right) = \left( {0,\frac{1}{{\sqrt 2 }}, - \frac{1}{{\sqrt 2 }}} \right)$
