Hello, tastylick!
$\displaystyle \text{Given: }\:w \:=\:\frac{9x+y}{3z-2y}$
$\displaystyle \text{Then: }\:\frac{\partial w}{\partial y} \:=\:\frac{1\cdot(3z-2y) - (\text{-}2)(9x+y)}{(3z-2y)^2}$
. . . . . . . . $\displaystyle =\:\frac{3z - 2y + 18x + 2y}{(3z-2y)^2}$
. . . . . . . . $\displaystyle =\:\frac{3z+18x}{(3z-2y)^2}$
The "total differential" of f(x, y, z) is $\displaystyle \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy+ \frac{\partial f}{\partial z}dz$
so your problem appears to be with the partial derivative of $\displaystyle \frac{9x+ y}{3z- 2y}$.
By the "quotient rule" that is $\displaystyle \frac{(1)(3z- 2y)- (-2)(9x+ y)}{(3z- 2y)^2}= \frac{3z+ 18x}{(3z- 2y)^2}$
You have left out the "x"!