Answer
$(a)$ $$z=\frac{k}{x^2y^3}$$
$(b)$ $$\frac{1}{72}$$
Work Step by Step
$(a)$
Applying inverse variation of $x^2$ and $y^3$ gives us the following:
$z=k\times\frac{1}{x^2}\times\frac{1}{y^3}=\frac{k}{x^2y^3}$
$(b)$ If we triple $x$ and double $y$, we will get:
$z=\frac{k}{(3x)^2(2y)^3}=\frac{k}{9x^2\times8y^3}=\frac{1}{72}\times\frac{k}{x^2y^3}$
It changes by a factor of $\frac{1}{72}$