Figure (a) shows a MOSFET cascode amplifying stage with a cascode active load. The small signal equivalent circuit is shown in figure (b) where Ro3 is the effective resistance looking into the drain of M3.

In cascode current mirror,

$R_{o3}=r_{03}+r_{04}(1+g_mr_{03})$

We can assume all transistors are matched so that the currents in all transistors are equal. Summing currents at D1, we have

$g_mV_{gs1}+\frac{(-V_{gs2})}{r_{o1}}=g_mV_{gs2}+\frac{V_0-(-V_{gs2})}{r_{02}}$

Summing currents at the output node, we find

$\frac{V_0}{R_{03}}+\frac{V_0-(-V_{gs2})}{r_{02}}+g_mV_{gs2}=0$

Eliminating Vgs2 from the two equations, noting that Vgs1 = Vin and assuming gm >> 1/ro, we find the small signal voltage gain is

$A_v=\frac{V_0}{V-1}=\frac{-g_m^2}{\frac{g_m}{R_{03}}+\frac{1}{r_{01}r_{02}}}$

The resistance Ro3 is approximately $R_{03}\cong g_{m1}r_{03}r_{04}$ so the gain can be written as

$A_v=\frac{-g_m^2}{\frac{1}{r_{03}r_{04}}+\frac{1}{r_{01}r_{02}}}$