Answer
$$
\begin{aligned}
d w= &\frac{10 x}{z+1} d x+\frac{2 y}{z+1} d y+\frac{-5 x^{2}-y^{2}}{(z+1)^{2}} d z\\
&= \frac{-20}{2}(0.02)+\frac{2}{2}(-0.03) +\frac{[-5(4)-1](0.02)}{(2)^{2}} \\
&=-0.2-0.03-\frac{21}{4}(0.02) \\
&=-0.335
\end{aligned}
$$
Work Step by Step
$$
w=\frac{5 x^{2}+y^{2}}{z+1}
$$
First find
$$
\begin{aligned}
f_{x}(x, y) &=\frac{(z+1) 10 x-\left(5 x^{2}+y^{2}\right)(0)}{(z+1)^{2}} \\
&=\frac{10 x}{z+1} \\
f_{y}(x, y)&=\frac{(z+1)(2 y)-\left(5 x^{2}+y^{2}\right)(0)}{(z+1)^{2}} \\
&=\frac{2 y}{z+1}\\
f_{z}(x, y)&=\frac{(z+1)(0)-\left(5 x^{2}+y^{2}\right)(1)}{(z+1)^{2}} \\
&=\frac{-5 x^{2}-y^{2}}{(z+1)^{2}}
\end{aligned}
$$
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=1, d x=0.02, d y=-0.03, d z=0.02 $ , we have :
$$
\begin{aligned}
d w= &\frac{10 x}{z+1} d x+\frac{2 y}{z+1} d y+\frac{-5 x^{2}-y^{2}}{(z+1)^{2}} d z\\
&= \frac{-20}{2}(0.02)+\frac{2}{2}(-0.03) +\frac{[-5(4)-1](0.02)}{(2)^{2}} \\
&=-0.2-0.03-\frac{21}{4}(0.02) \\
&=-0.335
\end{aligned}
$$
