How do I derive \$(i_x)/(V_A-V_B)\$ in this circuit?
Although your answer is almost right (except for the sign), I'm not sure your technique is. KCL doesn't directly yield \$g_mv_A-g_mv_B-2i_x=0\$. Further, \$v_A\$ and \$v_B\$ aren't base-emitter voltages as you've drawn, but are instead ground-referred.

Let's define a dummy voltage \$v_x\$, which is the voltage at the emitters. The pi model gives us \$i_{T_1}=g_m(v_A-v_x)\$, and \$i_{T_2}=g_m(v_B-v_x)\$. KCL at the emitter node forces \$i_{T_1}+i_{T_2}=0\$, thus \$g_m(v_A-v_x)+g_m(v_B-v_x)=0\$ and, therefore,
$$v_x=\frac{v_A+v_B}{2}$$
In other words, \$v_x\$ is the average of \$v_A\$ and \$v_B\$. Knowing that \$i_x\$ is defined to be \$i_{T_2}\$,
$$i_x=g_m(v_B-v_x)$$
$$i_x=g_m\left(v_B-\frac{v_A+v_B}{2}\right)$$
$$\therefore i_x=g_m\frac{v_B-v_A}{2}$$



Personally, I prefer using the T (tee) model for just about everything, and here's why.

If we define two "emitter resistors" strung between \$v_A\$ and \$v_B\$, each of value \$r_e=1/g_m\$ like so:







simulate this circuit &ndash; Schematic created using CircuitLab

then we know that the current that flows from \$v_B\$ to \$v_A\$ is simply that of two series resistors of value \$r_e\$,
$$i_x=\frac{v_B-v_A}{2r_e}=g_m\frac{v_B-v_A}{2}$$
Notice that I snuck in a claim that the current flowing from \$v_B\$ to \$v_A\$ is \$i_x\$. This is because, in the T model, [virtually] all the current that flows through the emitter resistors \$r_e\$ is supplied by a current source connected to the transistors' collectors. Thus, the current flowing from \$v_B\$ to \$v_A\$ is \$i_x\$.

Notice that we can also see, by inspection, that \$v_x\$ is \$(v_A+v_B)/2\$. This is an added bonus of the T model.