Answer
$$
\begin{aligned}
d w=&\left(\ln y+\ln z-\frac{y}{x}\right) d x \\
&+\left(\frac{x}{y}-\ln x+\ln z\right) d y \\
&+\left(\frac{x}{z}+\frac{y}{z}\right) d z \\
=&\left(\ln 1+\ln 4-\frac{1}{2}\right)(0.03) \\
&+\left(\frac{2}{1}-\ln 2+\ln 4\right)(0.02) \\
&+\left(\frac{2}{4}+\frac{1}{4}\right)(-0.01) \\
\approx & 0.0730
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
w=& x \ln (y z)-y \ln \left(\frac{x}{z}\right) \\
=& x(\ln y+\ln z)-y(\ln x-\ln z)
\end{aligned}
$$
First find
$$
\begin{aligned}
f_{x}(x, y, z) &=\left(\ln y+\ln z-\frac{y}{x}\right) \\
f_{y}(x, y,z)&=\left(\frac{x}{y}-\ln x+\ln z\right) \\
f_{z}(x, y,z)&=\left(\frac{x}{z}+\frac{y}{z}\right)
\end{aligned}
$$
Here
$$
d w= f_{x}(x, y, z) d x+f_{y}(x, y, z) d y+f_{z}(x, y, z) d z
$$
Substitute the given values, $x= 2, y=1, z=4, d x=0.03, d y=0.02 ,d z=-0.01 $ , we have :
$$
\begin{aligned}
d w=&\left(\ln y+\ln z-\frac{y}{x}\right) d x \\
&+\left(\frac{x}{y}-\ln x+\ln z\right) d y \\
&+\left(\frac{x}{z}+\frac{y}{z}\right) d z \\
=&\left(\ln 1+\ln 4-\frac{1}{2}\right)(0.03) \\
&+\left(\frac{2}{1}-\ln 2+\ln 4\right)(0.02) \\
&+\left(\frac{2}{4}+\frac{1}{4}\right)(-0.01) \\
\approx & 0.0730
\end{aligned}
$$
