Let $f$ be a function for which $f\left(\dfrac{x}{3}\right) = x^2 + x + 1$. Find the sum of all values of $z$ for which $f(3z) = 7$.
\[\text {(A)}\ -1/3 \qquad \text {(B)}\ -1/9 \qquad \text {(C)}\ 0 \qquad \text {(D)}\ 5/9 \qquad \text {(E)}\ 5/3\]
Let $y = \frac{x}{3}$; then $f(y) = (3y)^2 + 3y + 1 = 9y^2 + 3y+1$. Thus $f(3z)-7=81z^2+9z-6=3(9z-2)(3z+1)=0$, and $z = -\frac{1}{3}, \frac{2}{9}$. These sum up to $\boxed{\textbf{(B) }-\frac19}$.