Step 1: Match the equation to the general second degree form.
Compare $9x^2+axy+4y^2+6x+by-3=0$ with $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$. This gives $A=9$, $2H=a$, $B=4$, $2G=6$, $2F=b$, $C=-3$.
Step 2: Apply the condition for parallel lines.
A pair of straight lines is parallel exactly when the second degree part is a perfect square, i.e. $H^2 = AB$.
Step 3: Solve for $a$.
Since $H=\frac{a}{2}$, we need $\left(\frac{a}{2}\right)^2 = 9\times 4 = 36$, so $\frac{a^2}{4}=36$, giving $a^2 = 144$ and $a = \pm 12$.
Step 4: Factor the quadratic part.
Taking $a=12$, the second degree terms become $9x^2+12xy+4y^2 = (3x+2y)^2$. So the two parallel lines are of the form $3x+2y = k_1$ and $3x+2y=k_2$.
Step 5: Use the linear and constant terms to fix $b$.
The whole expression should factor as $(3x+2y+p)(3x+2y+q)$. Expanding, the $x$ term gives $3(p+q)=6$, so $p+q=2$; the constant gives $pq=-3$. Solving, $p=3,\,q=-1$ (or vice versa).
Step 6: Read off $b$.
The $y$ term is $2(p+q) = 2(2) = 4$, which equals $b$. So $a=12,\,b=4$.
\[ \boxed{a=12,\; b=4} \]