2.215. $\left((z-3)(z+3)^{-1}-\frac{(z+3)^{3 / 2}}{\sqrt{\left(z^{2}-9\right)(z-3)}}\right) \cdot \frac{\frac{1}{3}-\frac{z}{18}-\frac{1}{2 z}}{(z+3)^{-1}}$.
Solution.

Domain of definition: $\left\{\begin{array}{l}z \neq 0, \\ z>-3, \\ z \neq 3\end{array}\right.$

$$
\begin{aligned}
& \left((z-3)(z+3)^{-1}-\frac{(z+3)^{3 / 2}}{\sqrt{\left(z^{2}-9\right)(z-3)}}\right) \cdot \frac{\frac{1}{3}-\frac{z}{18}-\frac{1}{2 z}}{(z+3)^{-1}}= \\
& =\left(\frac{z-3}{z+3}-\frac{\sqrt{(z+3)^{3}}}{\sqrt{(z-3)^{2}(z+3)}}\right) \cdot \frac{\frac{6 z-z^{2}-9}{18 z}}{\frac{1}{z+3}}=\left(\frac{z-3}{z+3}-\frac{\sqrt{(z+3)^{2}}}{\sqrt{(z-3)^{2}}}\right) \times \\
& \times \frac{-\left(z^{2}-6 z+9\right)}{18 z}: \frac{1}{z+3}=\left(\frac{z-3}{z+3}-\frac{z+3}{|z-3|}\right) \cdot \frac{-(z-3)^{2}(z+3)}{18 z}
\end{aligned}
$$

By resolving the modulus with respect to the domain of definition, we get two cases:

$$
\begin{aligned}
& \text { i) }\left\{\begin{array}{l}
z \in(-3 ; 0) \cup(0 ; 3) \\
\left(\frac{z-3}{z+3}+\frac{z+3}{z-3}\right) \cdot \frac{-(z-3)^{2}(z+3)}{18 z}=\frac{(z-3)^{2}+(z+3)^{2}}{(z+3)(z-3)} \cdot \frac{-(z-3)^{2}(z+3)}{18 z}=
\end{array}\right. \\
& =-\frac{2\left(z^{2}+9\right)(z-3)}{18 z}=-\frac{\left(z^{2}+9\right)(z-3)}{9 z}=\frac{\left(z^{2}+9\right)(3-z)}{9 z} ; \\
& \text { 2) }\left\{\begin{array}{l}
z \in(3 ; \infty) \\
\left(\frac{z-3}{z+3}-\frac{z+3}{z-3}\right) \cdot \frac{-(z-3)^{2}(z+3)}{18 z}=\frac{(z-3)^{2}-(z+3)^{2}}{(z+3)(z-3)} \cdot \frac{-(z-3)^{2}(z+3)}{18 z}=
\end{array}\right. \\
& =\frac{12 z(z-3)}{18 z}=\frac{2(z-3)}{3} \text {. }
\end{aligned}
$$

Answer: $\frac{\left(z^{2}+9\right)(3-z)}{9 z}$, if $z \in(-3 ; 0) \cup(0 ; 3) ; \frac{2(z-3)}{3}$, if $z \in(3 ; \infty)$.