<s>[INST] Computing Young's modulus of an ideal elastic substance using it's thermodinamic equation of state 
The equation of state of an ideal elastic substance is:
\begin{equation}
    \mathcal{F} = KT \left[\left(\frac{L}{L_0}\right) - \left(\frac{L}{L_0}\right)^{-2}\right]
  \tag{1}
  \end{equation}
Where $K$ is a constant and $L_0$ (the value of $L$ at zero tension)
is a function of temperature only ($L_0(T)$).
Show that the isothermal Young's modulus is given by
\begin{equation}
    Y = \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2}
    \tag{2}
 \end{equation}
Exersice 2.7.a of: Heat and Thermodynamics 7th Revised edition. Mark W. Zemansky; Richard H. Dittman

If:
\begin{equation}
  Y = \frac{L}{A}\left(\frac{\partial \mathcal{F}}{\partial L}\right)_T
\tag{3}
\end{equation}
Can perform the derivative of equation 3 with equation 1 as:
\begin{equation}
  \begin{aligned}
    \mathcal{F} & = KT \left[\left(\frac{L}{L_0}\right)-\left(\frac{L}{L_0}\right)^{-2}\right] \\
                & =  \frac{KTL}{L_0} - \frac{KTL_0^2}{L^2}
  \end{aligned}
\tag{4}
\end{equation}
substituting 4 in 3, solvig the derivarives and reducing terms:
\begin{equation}
  \begin{aligned}
    Y & = \frac{L}{A} \left(\frac{\partial}{\partial L}\right)_T
    \left[\frac{KTL}{L_0} - \frac{KTL_0^2}{L^2}    \right]                                                                                                      \\
      & = \frac{L}{A} \left[ \left(\frac{\partial}{\partial L} \frac{KTL}{L_0}\right)_T - \left(\frac{\partial}{\partial L}\frac{KTL_0^2}{L^2}\right)_T \right] \\
      & = \frac{L}{A} \left[\frac{KT}{L_0} + \frac{2KL_0^2T}{L^3}      \right]                                                                                  \\                                                                                  \\
      & = \frac{L}{A} \left[\frac{KTL^3 + 2KL_0^3T}{L^3 L_0}      \right]                                                                                       \\
      & = \frac{KTL^3 + 2KL_0^3T}{AL^2 L_0}                                                                                                                     \\
      & = KT\frac{L^3 +2L_0^3}{AL^2 L_0}                                                                                                                        \\
      & = KT\left[ \frac{L}{AL_0} + \frac{2L_0^2}{AL^2}      \right]                                                                                            \\
      & = \frac{KTL}{AL_0} + \frac{2KTL_0^2}{AL^2}                                                                                                              \\  \end{aligned}
\end{equation}
Which:
\begin{equation}
\boxed{
   \frac{KTL}{AL_0} + \frac{2KTL_0^2}{AL^2}  \neq \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2}
}
\end{equation}
Then just by solving the derivative $\partial \mathcal{F}/\partial L$ of eq.3 in eq 1. leadsme to a path where I miss the term $\mathcal{F}/A$ in eq. 2. That's where I think I'm missing something in the theory, which is what I'm looking for.
Even if I can demostrate that:
\begin{equation}
  \mathcal{F} = \frac{KTL}{L_0}
\end{equation}
then:
\begin{equation}
\boxed{
   \frac{\mathcal{F}}{A} + \frac{2KTL_0^2}{AL^2}  \neq \frac{\mathcal{F}}{A} + \frac{3KTL_0^2}{AL^2}
}
\end{equation}
 [/INST] Chemomechanics answered your question, I am merely doing the algebra:
You say
$$
\boxed{
\frac{k T L}{A L_0} + \frac{2 k T L_0^2}{A L^2}
\neq
\frac{\mathcal F}{A}+\frac{3 k T L_0^2}{A L^2}
}
$$
but this is not true: From your Eq (1)
$$
   \frac{\mathcal F}{A} =\frac{kT L}{AL_0}-\frac{k T L_0^2}{L^2}
$$
Then
$$\boxed{\boxed{
  \frac{\mathcal F}{A} + \frac{3 k T L_0^2}{A L^2} =\frac{kT L}{AL_0}+ \frac{2 k T L_0^2}{A L^2} 
}}
$$
</s>