The cumulative distribution function (CDF) is the anti-derivative of your probability density function (PDF). So, you need to find the indefinite integral of your density. Only if you are given the CDF, you can take its first derivative in order to obtain the PDF. However, your proposed function is not a density function because a density has the property that it integrates to one. This does not hold here:
$\int_0^3 9t^2 dt = [3t^3]_0^3 = 3 \cdot 3^3 = 3^4 = 81.$
So, this is not a density function and you either need to adjust the boundaries or scale it by $\frac{1}{81}$. An adjustment of the boundaries could be achieved by changing the upper bound $3$ to $\sqrt[3]{\frac{1}{3}}$.
EDIT: The OP adjusted the formula after my remark above to a PDF
Now, your function is a density. See, it integrates to one:
$\int_0^3 \frac{1}{9}t^2 dt = [\frac{1}{27}t^3]_0^3 = \frac{1}{27}3^3 = \frac{1}{27}27= 1.$
For getting the CDF $F(t)$, you need to find the indefinite integral:
$F(t) = \int \frac{1}{9}t^2 dt = \begin{cases}
\frac{1}{27}t^3 & \text{if } t \in (0,3)\\
0 & \text{otherwise}
\end{cases}$