Why doesn't the voltage divider equation work with this circuit?
You can use the Thevenin equivalent for voltage dividers, and particularly with regard to the \$R_3\$ and \$R_4\$ combination with \$V_O\$, in an intuitive way:

$$\begin{align*}
R_{TH}&amp;=\frac{R_3\cdot R_4}{R_3+R_4}\\\\
V_{TH}&amp;=V_O\cdot\frac{R_3}{R_3+R_4}\\\\
\end{align*}$$

Now, \$V_{TH}\$ is in series with \$R_{TH}\$ and \$R_2\$ before reaching the virtual ground \$V_-\$ node. So set up the equality for the currents and proceed:

$$\begin{align*}
\frac{V_{TH}}{R_{TH}+R_2}&amp;=-\frac{V_I}{R_1}\\\\
\frac{V_O\cdot\frac{R_3}{R_3+R_4}}{\frac{R_3\cdot R_4}{R_3+R_4}+R_2}&amp;=-\frac{V_I}{R_1}\\\\
\therefore \frac{V_O}{V_I}&amp;=-\frac{\frac{R_3\cdot R_4}{R_3+R_4}+R_2}{R_1\cdot\frac{R_3}{R_3+R_4}}\\\\
&amp;=-\frac{R_3\cdot R_4+R_2\cdot R_3+R_2\cdot R_4}{R_1\cdot R_3}\\\\
&amp;=-\left[\frac{R_4}{R_1}+\frac{R_2}{R_1}+\frac{R_2\cdot R_4}{R_1\cdot R_3}\right]\\\\
&amp;=-\frac{1}{R_1}\cdot\left[R_4+R_2+\frac{R_2\cdot R_4}{R_3}\right]\\\\
&amp;=-\frac{R_2}{R_1}\cdot\left[1+\frac{R_4}{R_2}+\frac{R_4}{R_3}\right]
\end{align*}$$

So, this means that you can use Thevenin, just fine. As always, laws just work. It's just about following the implications accurately as you proceed through.

Please note that I didn't say that \$V_X=V_{TH}\$. \$V_X\$ is a physical node in your circuit. \$V_{TH}\$ is a virtual Thevenin node with a series resistance, \$R_{TH}\$, that leads to, but is not the same as, \$V_X\$. If you want \$V_X\$, then:

$$V_X=V_{TH}\cdot\frac{R_2}{R_{TH}+R_2}$$

Note that this isn't what you wrote. That's probably because you misunderstood and imagined that the Thevenin voltage was actually the same as \$V_X\$. However, it's not the same thing.