I want to use exterior calculus and Legendre transformation consideration to generate Maxwell relations corresponding to the following wedge product
$$dT \wedge dN$$
given $$dE=T dS-P dV+\mu dN$$ where $E$ is energy; $\dfrac{\partial E}{\partial S}$ is temperature, T; $-\dfrac{\partial E}{\partial V}$ is pressure, P; $\dfrac{\partial E}{\partial N}$ is the chemical potential $\mu$. $V$ is volume, $S$ is entropy and $N$ is number of molecules.
I already have the answer to this question, which would be $$\dfrac{\partial \mu}{\partial T}=-\dfrac{\partial S}{\partial N}$$
but I am not sure how did we come to this solution... the question has many parts, but I just need to understand this part so I can do the rest on my own.
The usual deduction from $d(dE)=0$ would be things like $\dfrac{\partial T}{\partial N} = \dfrac{\partial\mu}{\partial S}$ (coming from the coefficient of $dN\wedge dS$. But they are doing something a bit trickier.
However, what's missing from the equation is the usual thermodynamic indication of what the independent variables are in each case. Is $\dfrac{\partial\mu}{\partial T}$ really $\left(\dfrac{\partial\mu}{\partial T}\right)_{N,P,V}$ or something like that?
Again, we take $0=d(dE)=dT\wedge dS - dP\wedge dV + d\mu\wedge dN$. Then the coefficient of $dT\wedge dN$ (assuming $P,V$ don't depend on $T,N$) will have to be $0$, and this coefficient is $\dfrac{\partial S}{\partial N} + \dfrac{\partial\mu}{\partial T}$. Setting this equal to $0$ gives your result.