Using a diode-connected MOSFETs as a voltage divider

You're mixing up / combining small signal behavior and DC behavior.

The \$1/g_m\$ "resistance" of a MOSFET (it applies to JFETs and BJTs as well) is small signal behavior. So if you have a small signal (for example, a sinewave of 10 mV) at \$V_{in}\$ and

\$g_{m,M1}\$ = \$g_{m,M2}\$

then at \$V_{out}\$ you get a half of the input signal (so a sinewave of 5 mV).

If you change \$g_{m,M1}\$ and \$g_{m,M2}\$ you will indeed find that output voltage \$V_{out}\$will follow the resistive divider formula that you show.

However the behavior for DC voltages does not work in the same way. Then the formula \$I_D = \mu C_{OX}\frac W L (V_{GS} - V_{TH})\$ applies. If you would use two identical transistors and make \$V_{in}\$ = 5 V you would get \$V_{out}\$ = 2.5 V (I'm ignoring the body effect here!).

However, if you would make the bottom transistor M2 2 x larger (you could just put two identical MOSFETs in parallel) then you would not get \$V_{out}\$ = \$\frac 1 3\ V_{in}\$.

That is so because the threshold voltage \$V_{TH}\$ remains the same for M1 and (the double-size) M2. So although \$(V_{GS} - V_{TH})\$ of M2 will be half of the \$I_D = \mu C_{OX}\frac W L (V_{GS} - V_{TH})\$ of M1, this is not true for the threshold voltage \$V_{TH}\$.

Look at it this way, if we had transistors with a threshold voltage \$V_{TH}\$ = 0 (zero) then we would get:

\$V_{out}\$ = \$\frac 1 3\ V_{in}\$.

But the threshold voltage \$V_{TH}\$ is not zero so that "messes up" the voltage division ratio for DC voltages.

If one neglects the current that may flow through the \$V_{out}\$ connection, then both transistors will have the same current flowing through them. If the transistors have identical characteristics, then each transistor will have the same voltage \$V_{DS}=V_{GS}\$ across it's terminals, both will have the same \$g_m\$. This is true regardless of the applied voltage. If the current you draw from the output is small enough (relative to the current through the transistors), your output will be approximately 0.5 times your input. But that assumes your transistors are perfectly matched and the current drawn from the output is small enough.

When I change my Vin to 1V, my 1/gm should change, hence my resistance should change and now I will no longer get my 0.5 ratio.

The current through each transistor will change, but they should change equally. (Again, assuming we can neglect the current lost through \$V_{out}\$ connection. The problem however, is that you are using mosfets, which are (generally) enhancement mode devices. They do not conduct appreciably unless the gate-source voltage is at or above a threshold voltage. To get the transitors to conduct with 1V across both, will require a threshold voltage of 0.5V or less.

I don't really really know the details of your application, so the following suggestion may not suit your needs, but you may have better luck with very low voltages if you use identical JFETs configured with their source and gate connected. These will continue to conduct all the way down to 0V.

Or could you simply use resistors?

Edit: The reason why the voltage drop between the two transistors will tend toward equality is as follows. (We will assume there is no output current, at least for now.)

Hopefully you can see that if the transistors have identical characteristics, then if the current through each is equal, the voltages across them will be equal, and vice versa.