Finding Rout of differential pair with active current mirror (from Razavi's "Design of Analog CMOS Integrated Circuits 1st Edition")
\$I_{d4} = I_{d3} + \frac{V_x}{R_{o4}}\$ because M3:M4 is a current mirror which would enforce \$I_{d3} = I_{d4}\$, except for the \$R_{o4}\$, which is the effect of channel length modulation. 

None of the small-signal current flows through Iss, so \$I_{xy} = I_{d3}\$. Then, \$I_x = I_{xy} + I_{d4}\$
so \$I_x = I_{xy} + I_{xy} + V_x/R_{o4}\$.