Find $z^{\text{'}}(t),$ where $z=\frac{1}{x}\frac{1}{y},x=t^{2}4t,$ and $y=t^3-6,$ in the following ways.

a$.$ Replace $x$ and $y$ to write $z$ as a function of $t$ and differentiate.

b$.$ Use the Chain Rule.

$($Type an expression using $t$ as the variable.$)$

$\frac{\text{d}\text{e}\text{l}\text{z}}{\text{d}\text{e}\text{l}\text{y}}=-\frac{1}{y^2}$

$($Type an expression using $x$ and $y$ as the variables.$)$

$\frac{dy}{dt}=3t^2$

$($Type an expression using $t$ as the variable.$)$

Using either method, $z^{\text{'}}(t)=-\frac{(2t4)}{(t^{2}4t{)}^2}-\frac{(3t^2)}{(t^{3}6{)}^2}$

$($Type an expression using t as the variable.$)$