jmc

Since the graph of $f$ contains the point $(-3,a)$, we know that $$a=f(-3)=3(-3)+b=b-9.$$Since the graph of $f^{-1}$ also contains this point we know that $f^{-1}(-3)=a$ or $-3=f(a)$. Therefore $$-3=f(a)=3a+b.$$Substitution for $a$ gives $$-3=3(b-9)+b=4b-27.$$Therefore $b=\frac{1}{4}(27-3)=6$. This forces $$a=b-9=6-9=\boxed{-3}.$$One could also recall that the graph of $f$ is a line and the graph of $f^{-1}$ is that line reflected through $y=x$. Since the slopes of these lines are not 1, the lines both intersect $y=x$ at a single point and that point is also the point of intersection of the graphs of $f$ and $f^{-1}$. Therefore the intersection point must be $(-3,-3)$, giving $a=\boxed{-3}$.