Z-Transform of $x(n) = 3^n$
The signal $x[n]=3^n$ for $n\le -3$ (and zero otherwise, I assume) can be written as

$$x[n]=3^nu[-n-3]\tag{1}$$

To compute its $\mathcal{Z}$-transform you simply need to use the formula:

$$\begin{align}X(z)&amp;=\sum_{n=-\infty}^{\infty}x[n]z^{-n}\\&amp;=\sum_{n=-\infty}^{-3}3^nz^{-n}\\&amp;=\sum_{n=-\infty}^03^{(n-3)}z^{-(n-3)}\\&amp;=3^{-3}z^{3}\sum_{n=0}^{\infty}\left(\frac{z}{3}\right)^n\\&amp;=3^{-3}z^{3}\frac{1}{1-\frac13 z},\quad |z|&lt;3\tag{2}\end{align}$$

It's important to specify the region of convergence ($|z|&lt;3$), otherwise the $\mathcal{Z}$-transform does not uniquely identify the corresponding sequence $x[n]$.