Answer
$${\bf{N}}\left( 0 \right) = \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{{\sqrt 2 }}{2}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \sqrt 2 t{\bf{i}} + {e^t}{\bf{j}} + {e^{ - t}}{\bf{k}},{\text{ }}t = 0 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{\sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}}}}{{\left\| {\sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}}} \right\|}} = \frac{{\sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}}}}{{\sqrt {2 + {e^{2t}} - {e^{ - 2t}}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{\sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}}}}{{\sqrt {{{\left( {{e^t} + {e^{ - t}}} \right)}^2}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{\sqrt 2 {\bf{i}} + {e^t}{\bf{j}} - {e^{ - t}}{\bf{k}}}}{{{e^t} + {e^{ - t}}}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{{\sqrt 2 }}{{{e^t} + {e^{ - t}}}}} \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\frac{{{e^t}}}{{{e^t} + {e^{ - t}}}}} \right]{\bf{j}} - \frac{d}{{dt}}\left[ {\frac{{{e^{ - t}}}}{{{e^t} + {e^{ - t}}}}} \right]{\bf{k}} \cr
& {\text{Making the operations by hand and replacing }}\left( {{\text{or use a CAS}}} \right) \cr
& {\bf{T}}'\left( t \right) = - \frac{{\sqrt 2 \left( {{e^t} - {e^{ - t}}} \right)}}{{{{\left( {{e^t} + {e^{ - t}}} \right)}^2}}}{\bf{i}} + \frac{2}{{{{\left( {{e^t} + {e^{ - t}}} \right)}^2}}}{\bf{j}} + \frac{2}{{{{\left( {{e^t} + {e^{ - t}}} \right)}^2}}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 0 \cr
& {\bf{T}}'\left( 0 \right) = - \frac{{\sqrt 2 \left( {{e^0} - {e^{ - 0}}} \right)}}{{{{\left( {{e^0} + {e^{ - 0}}} \right)}^2}}}{\bf{i}} + \frac{2}{{{{\left( {{e^0} + {e^{ - 0}}} \right)}^2}}}{\bf{j}} + \frac{2}{{{{\left( {{e^0} + {e^{ - 0}}} \right)}^2}}}{\bf{k}} \cr
& {\bf{T}}'\left( 0 \right) = 0{\bf{i}} + \frac{2}{{{{\left( 2 \right)}^2}}}{\bf{j}} + \frac{2}{{{{\left( 2 \right)}^2}}}{\bf{k}} \cr
& {\bf{T}}'\left( 0 \right) = \frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 0 \cr
& {\bf{N}}\left( 0 \right) = \frac{{{\bf{T}}'\left( 0 \right)}}{{\left\| {{\bf{T}}'\left( 0 \right)} \right\|}} = \frac{{\frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}}}}{{\left\| {\frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}}} \right\|}} \cr
& {\bf{N}}\left( 0 \right) = \frac{{\frac{1}{2}{\bf{j}} + \frac{1}{2}{\bf{k}}}}{{\sqrt 2 /2}} \cr
& {\bf{N}}\left( 0 \right) = \frac{{\sqrt 2 }}{2}{\bf{j}} + \frac{{\sqrt 2 }}{2}{\bf{k}} \cr} $$
