Answer
\begin{aligned} d w &=\frac{1}{z-3 y} d x+\frac{3 x+z}{(z-3 y)^{2}} d y-\frac{x+y}{(z-3 y)^{2}} d z \end{aligned}
Work Step by Step
Given $$ w =\frac{x+y}{z-3 y}$$
Since $$dw=\frac{\partial w}{\partial x} dx+\frac{\partial w}{\partial y} dy+\frac{\partial w}{\partial z} dz,$$
$$\frac{\partial w}{\partial x} =\frac{(1)(z-3 y)-0}{(z-3 y)^2} =\frac{1}{(z-3 y)} ,$$
$$\frac{\partial w}{\partial y} =\frac{(1)(z-3 y)-(-3)(x+y)}{(z-3 y)^2} =\frac{ z-3 y +3x+3y)}{(z-3 y)^2} \\=\frac{ (z +3x )}{(z-3 y)^2}$$
and
$$\frac{\partial w}{\partial z} = \frac{0-(1)(x+y)}{(z-3 y)^2} =-\frac{(x+y)}{(z-3 y)^2} $$
then we get
\begin{aligned} d w &=\frac{1}{z-3 y} d x+\frac{3 x+z}{(z-3 y)^{2}} d y-\frac{x+y}{(z-3 y)^{2}} d z \end{aligned}
