Answer
$${\bf{N}}\left( {\frac{\pi }{4}} \right) = - {\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 3\cos 2t{\bf{i}} + 3\sin 2t{\bf{j}} + 3{\bf{k}},{\text{ }}t = \frac{\pi }{4} \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {3\cos 2t{\bf{i}} + 3\sin 2t{\bf{j}} + 3{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 6\sin 2t{\bf{i}} + 6\cos 2t{\bf{j}} \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{{ - 6\sin 2t{\bf{i}} + 6\cos 2t{\bf{j}}}}{{\left\| { - 6\sin 2t{\bf{i}} + 6\cos 2t{\bf{j}}} \right\|}} = \frac{{ - 6\sin 2t{\bf{i}} + 6\cos 2t{\bf{j}}}}{{\sqrt {36{{\sin }^2}2t + 36{{\cos }^2}2t} }} \cr
& {\bf{T}}\left( t \right) = \frac{{ - 6\sin 2t{\bf{i}} + 6\cos 2t{\bf{j}}}}{6} \cr
& {\bf{T}}\left( t \right) = - \sin 2t{\bf{i}} + \cos 2t{\bf{j}} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ { - \sin 2t{\bf{i}} + \cos 2t{\bf{j}}} \right] \cr
& {\bf{T}}'\left( t \right) = - 2\cos 2t{\bf{i}} - 2\sin 2t{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = \frac{\pi }{4} \cr
& {\bf{T}}'\left( {\frac{\pi }{4}} \right) = - 2\cos 2\left( {\frac{\pi }{4}} \right){\bf{i}} - 2\sin 2\left( {\frac{\pi }{4}} \right){\bf{j}} \cr
& {\bf{T}}'\left( {\frac{\pi }{4}} \right) = - 2{\bf{j}} \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 = \frac{\pi }{4} \cr
& {\bf{N}}\left( {\frac{\pi }{4}} \right) = \frac{{{\bf{T}}'\left( {\pi /4} \right)}}{{\left\| {{\bf{T}}'\left( {\pi /4} \right)} \right\|}} = \frac{{ - 2{\bf{j}}}}{{\left\| {2{\bf{j}}} \right\|}} \cr
& {\bf{N}}\left( {\frac{\pi }{4}} \right) = - {\bf{j}} \cr} $$
