Finding Maxwell relations 
I feel I am missing something about deriving Maxwell relations. I have read http://ocw.mit.edu/courses/physics/8-044-statistical-physics-i-spring-2013/readings-notes-slides/MIT8_044S13_notes.Max.pdf and as far as I can see if we have $dE=TdS+FdL$, the only maxwell relation will be
$$\frac{\partial F}{\partial S}=\frac{\partial T}{\partial L} $$
So how is the given maxwell relation in the picture found?
 A: As you said, the relation you have is
$$
dE = TdS + FdL
$$
So one maxwell relation is
$$
\left(\frac{\partial T}{\partial L}\right)_S = \left(\frac{\partial F}{\partial S}\right)_L
$$
Which is the one you have obtained. For get the other do the following transformation
$$
dE = TdS + FdL = d(ST) - SdT + FdL
$$
$$
\implies d(E-ST) = -SdT + FdL
$$
Since $-SdT + FdL$ is an exact differential, the cross derivatives coincides, so you can read the other maxwell relation
$$
-\left(\frac{\partial S}{\partial L} \right)_T = \left(\frac{\partial F}{\partial T} \right)_L 
$$
Which is the one on your textbook.
A: Specifically the result you have is $$\left(\frac{\partial F}{\partial S}\right)_L=\left(\frac{\partial T}{\partial L}\right)_S$$
So using the cyclic rule for partial derivatives we can write that 
$$ \left(\frac{\partial T}{\partial L}\right)_S=-\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial L}\right)_T$$
So if we sub this in and multiply both sides by $\left(\frac{\partial S}{\partial T}\right)_L$ we have 
$$\left(\frac{\partial F}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=-\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L \left(\frac{\partial S}{\partial L}\right)_T$$
and since 
$$\left(\frac{\partial F}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=\left(\frac{\partial F}{\partial T}\right)_L$$
and 
$$\left(\frac{\partial T}{\partial S}\right)_L \left(\frac{\partial S}{\partial T}\right)_L=1$$
you get the required result
$$\left(\frac{\partial F}{\partial T}\right)_L=-\left(\frac{\partial S}{\partial L}\right)_T$$
