How to apply chain rule to $\log$ function I'm doing my economics reading and I find this equation for elasticity of substitution:
$$\sigma \equiv \frac{F_K \cdot F_L}{F \cdot F_{KL}}$$ and then $$\frac{d(\log(F_K/F_L))}{d(\log(K/L))}=-\frac{1}{\sigma}.$$
How do I do the later derivative? By chain rule or anything else?
Many thanks. 

Take as a reference Chapter 5 in the book "Economists' Mathematical Manual" by Knut Sydsaeter, Arne Strøm, Peter Berck (where unfortunately no proof is given). 
As far as I understand $F$ is some production function with inputs $K$ and $L$ (capital and labor?). If $F(K,L)=c$ then $F_KdK+F_L dL=0$ which implies that $dK/dL=-F_L/F_K$. Therefore, with $F_{LK}=F_{KL}$, we get
$$\frac{d(\log(F_K/F_L))}{d(\log(K/L))}=
\frac{\frac{F_{KK}dK+F_{KL}dL}{F_K}-\frac{F_{LK}dK+F_{LL}dL}{F_L}}{\frac{dK}{K}-\frac{dL}{L}}=\frac{\frac{F_{KK}}{F_K^2}-\frac{2F_{KL}}{F_KF_L}+\frac{F_{LL}}{F_L^2}}{\frac{1}{KF_K}+\frac{1}{LF_L}}\tag{1}.$$
Now if $F(K,L)$ is homogeneous of degree 1 then 
$KF_K+LF_L=F$ and by taking the derivatives with respect to $K$ and $L$ we find
$$F_K+KF_{KK}+LF_{KL}=F_K\quad\text{and}\quad KF_{KL}+F_L+LF_{LL}=F_L.$$
Hence, $KF_{KK}=-LF_{KL}$, $LF_{LL}=-KF_{KL}$, and the right-hand side of (1) simplifies to
$$-\frac{FF_{KL}}{F_KF_L}=-\frac{1}{\sigma}.$$
