#### Answer

$a_1=-\dfrac{1}{27}$
$a_2=\dfrac{1}{9}$

#### Work Step by Step

To find the first and second terms, the value of the common ratio $r$ is needed.
Note that the next term of a geometric sequence can be found by multiplying the common ratio $r$ to the current term.
The known terms are:
$a_3=-\dfrac{1}{3}
\\a_6 = 9$
Note that using the third term as reference point, the sixth term can be found by multiplying the common ratio $r$ three times to the value of the third term.
Thus,
$a_6=a_3 \cdot r \cdot r \cdot r
\\a_6 = a_3 \cdot r^3$
Since $a_6=9$ and $a_3=-\dfrac{1}{3}$, substituting these values gives:
$a_6=a_3 \cdot r^3
\\9 = -\dfrac{1}{3} \cdot r^3
\\-3 \cdot 9 = -3 \cdot (-\dfrac{1}{3} \cdot r^3)
\\-27 = r^3
\\(-3)^3 = r^3$
Take the cube root of both sides to obtain:
$-3=r$
With $r=-3$, the second term can be computed by dividing $a_2$ by the common ratio to obtain:
$a_2 = \dfrac{a_3}{r}
\\a_2 = \dfrac{-\frac{1}{3}}{-3}
\\a_2=-\dfrac{1}{3} \cdot \dfrac{1}{-3}
\\a_2=\dfrac{1}{9}$
The first term can be found by dividing the second term by the common ratio:
$a_1 = \dfrac{a_2}{r}
\\a_1=\dfrac{\frac{1}{9}}{-3}
\\a_1=\dfrac{1}{9} \cdot \dfrac{1}{-3}
\\a_1=-\dfrac{1}{27}$