Answer
$$y=3x+3$$
Work Step by Step
Given
$$x=t^2-t,\ \ \ y=t^2+t+1 \ \ \ ;(0,3)$$
Since at $(0,3)$, we have
\begin{aligned}x& =t^2-t\\
0&=t(t-1) \end{aligned}
Then
$$ t=0,\ \ \ \ t=1 $$
Since $y=3$ at $t=1 $, then we reject $t=0$
\begin{aligned}
\frac{d y}{d t}=2t+1, \ \ \ \ \frac{d x}{d t}=2t-1\end{aligned}
Then
\begin{aligned}
\frac{d y}{d x}&=\frac{d y / d t}{d x / d t}\\
&=\frac{2t+1}{2t-1}
\end{aligned}
Hence the slope at $(0,3)$
\begin{aligned}
m&= 3
\end{aligned}
The tangent line equation given by
\begin{aligned}
\frac{y-y_1}{x-x_1}&=m\\
\frac{y-3}{x-0}&=3\\
y-3&= 3x\\
y&=3x+3
\end{aligned}
