Answer
$$ \mathbf{T}=\frac{3\sqrt{3}}{2\sqrt{7} } \mathbf{i} +\frac{1}{2\sqrt{7}} \mathbf{j} $$
Work Step by Step
Given $$\mathbf{r}= 6\sin t\mathbf{i} -2\cos t\mathbf{j} $$
Since
$$ \mathbf{r'}= 6\cos t\mathbf{i} +2\sin t\mathbf{j} $$
and $$ \| \mathbf{r'} \|=\sqrt{36 \cos^2 t+ 4\sin^2 t } =2 \sqrt{\sin^2 t+9\cos^2 t }$$
Then unit tangent vector given by
\begin{align*}
\mathbf{T}(t)&=\frac{\mathbf{r'}(t)}{\|\mathbf{r'}(t)\|}\\
&=\frac{ 6\cos t\mathbf{i} +2\sin t\mathbf{j}}{2 \sqrt{\sin^2 t+9\cos^2 t }}\\
&=\frac{ 3\cos t\mathbf{i} + \sin t\mathbf{j}}{ \sqrt{\sin^2 t+9\cos^2 t }}
\end{align*}
At $t=\pi/6$
$$ \mathbf{T}=\frac{3\sqrt{3}}{2\sqrt{7} } \mathbf{i} +\frac{1}{2\sqrt{7}} \mathbf{j} $$
