How to derive the molar fundamental relation We have the fundamental relation: 
$$dS = \frac{1}{T}dU+\frac{P}{T}dV-\frac{\mu}{T}dN$$
how do we get now the following:
$$ds = \frac{1}{T}du+\frac{P}{T}dv$$ where $s = \frac{S}{N}, u = \frac{U}{N}, v = \frac{V}{N}$.
I tried substitution but it got me nowhere..
You have  $$ds=\frac{du}{T}+\frac{P}{T}dv+\left(\frac{u}{T}-s+\frac{Pv}{T}-\frac{\mu}{T}\right)\frac{dN}{N}$$
But, $$\mu=u+Pv-Ts$$
That is, $\mu$ is the Gibbs free energy per mole.  So the term in parenthesis drops out.