Small-signal output resistance of MOS common-source stage with source degeneration
You are just measuring the total output resistance, in the absence of any stimulus, but the book says (and shows) that you are using \$\Delta V_{RS}\$ as the output of the \$\Delta V\$ input. Which means what you've written is just the denominator. You now have a resistive divider formed by the equivalent \$R_{in}=r_o\$ and \$R_{out}=R_S||\dfrac{1}{g_m+g_{mb}}\$, the latter being the one with \$\Delta V_{RS}\$ across: \$\Delta V_{RS}=\Delta V\dfrac{R_{out}}{R_{in}+R_{out}}=...\$ (I'll let you fill in the blanks; hint -- look at eq. 3.69).

$$\begin{align}
R_{eq}&amp;=R_S||\dfrac{1}{g_m+g_{mb}} \\
&amp;=\dfrac{R_S}{(g_m+g_{mb})R_S+1}\tag{1} \\
\dfrac{\Delta V_{RS}}{\Delta V}&amp;=\dfrac{R_{eq}}{R_{eq}+r_o} \\
&amp;=\dfrac{R_S}{(1+(g_m+g_{mb})R_S)r_o+R_S}\tag{2} \\
\Delta V_{RS}&amp;=\Delta V\dfrac{R_S}{(1+(g_m+g_{mb})R_S)r_o+R_S}\tag{3}
\end{align}$$
This leads to 3.70 and onwards.