Answer
$81y^4-108y^3+54y^2z^2-12yz^3+z^4$
Work Step by Step
Using the coefficients in the $r=5$ row of the Pascal's triangle, $\left\{
1,4,6,4,1
\right\}$, and the pattern in expanding binomials, then $
(3y-z)^4
$ expands to
\begin{array}{l}
(3y)^4(-z)^0+4(3y)^3(-z)^1+6(3y)^2(-z)^2+4(3y)^1(-z)^3+(3y)^0(-z)^4
\\\\=
81y^4-108y^3+54y^2z^2-12yz^3+z^4
.\end{array}