Answer
$(a)$ $$z=k\frac{x^2}{y^4}$$
$(b)$ $$\frac{9}{16}$$
Work Step by Step
$(a)$
Applying direct variation of $x^2$ and inverse variation of $y^4$ gives us the following:
$z=k\times x^2 \times \frac{1}{y^4}=k\frac{x^2}{y^4}$
$(b)$ If we triple $x$ and double $y$, we will get:
$z=k\frac{(3x)^2}{(2y)^4}=k\frac{9x^2}{16y^4}=\frac{9}{16}k\frac{x^2}{y^4}$
It changes by a factor of $\frac{9}{16}$