Suppose that $F(\cdot)$ has CRS in $K$ and $L$, the elasticity of Substitution is $\sigma_{K L} \equiv F_{L} F_{K} / F F_{L K}$.
I once derived this equation but I remember that it takes me quite long time and the process is quite tedious. I wonder is there any easy way to derive this or remember it intuitively?
I am not sure if it is intuitive but this is because because CRS function is homogenous of degree 1.
Full derivation:
First, general formula for any arbitrary elasticity of substitution between $L$ and $K$* is given by (see Sydsæter et al. EMEA pp 430):
$$\sigma_{L,K} = \frac{-F_K'F_L'(xF_K'+ yF_L')}{xy \left( (F_L')^2F_{KK}^{''} - 2 F_K'F_L'F_{KL}'' + (F_K')^2F_{LL}''\right)}, \text{ for: } F(K,L)=c \tag{1}$$
Where c is arbitrary constant. This is our starting point.
Now by definition property of CRS production function is that they are homogenous of degree 1 (since by definition we have CRS when $F(tK,tL)= tF(K,L)$).
If $F$ is homogenous of degree 1, then the numerator of (1) will be $= −F_K'F_L'F$.
This is because of the Euler's theorem which tells us that if:
$$f(x,y) \text{ is homogenous of degree k} \implies xf_x'(x,y) + yf_y'(x,y)= kf(x,y)$$
Next Euler theorem also implies that (assuming $f$ is twice continuously differentiable) that:
$$xf_{xx}''(x,y) + yf_{yx}''(x,y)= (k-1)f_x'(x,y)$$ $$xf_{yx}''(x,y) + yf_{yy}''(x,y)= (k-1)f_y'(x,y)$$
The above implies that in our case:
$$K F_{KK}'' = - LF_{KL}'' $$ and that $$L F_{LL}'' = - K F_{LK}'' = - K F_{KL}'' $$. Hence the denominator will be given by:
$$ -F{KL}''\left( L^2 (F_L')^2 + 2KLF_K'F_L'+ K^2(F_K')^2\right)= - F_{KL}''(KF_K'+ L F_L')^2 = - F_{KL}''F^2$$
We use Euler theorem above again.
Now finally we are done:
$$\sigma_{LK} = \frac{−F_K'F_L'F}{- F_{KL}''F^2} =\frac{F_K'F_L'}{ F_{KL}''F} $$
I personally do not think the result above is very intuitive (if it is so its intuition eludes me), but it is a consequence of CRS function being homogenous of degree 1 and this result actually applies for any elasticity of substitution between two variables of arbitrary function that is homogenous of degree 1 and twice continuously differentiable.
* or $y$ and $x$ for that matter this generalizes to any elasticity of substitution problem
1muflon1's answer is entirely correct.
Let me give another alternative derivation, which might be a little bit easier to remember, although probably not more intuitive.
Let $k = K/L$ be the capital to labour ratio. Then we can define $f(k) = F(K/L, 1)$ to be the output per unit of labour. then by the CRS assumption we have: $$ F(K,L) = L f(K/L) = L f(k). $$ Taking derivatives with respect to $K$ and $L$ gives: $$ \begin{align*} &F_K = L f'(k) \frac{1}{L} = f'(k),\\ &F_L = f(k) + L f'(k) \left(-\frac{K}{L^2}\right) = f(k) - k f'(k). \end{align*} $$ Finally: $$ \begin{align*} &F_{K,L} = f''(k)\left(-\frac{K}{L^2}\right) = -kf''(k) \frac{1}{L},\\ \to &F_{K,L} F = -k f''(k) f(k). \end{align*} $$ Let's now have a look at the IES: $$ \sigma_{L,K} = -\frac{\partial \ln(F_K/F_L)}{\partial \ln (K/L)} = \frac{\partial \ln \left(\frac{f'(k)}{f(k) - k f'(k)}\right) }{\partial \ln k} $$ Now take derivative of both numerator and denominator with respect to $k$: $$ \begin{align*} \sigma_{L,K} &= -k \frac{f - k f'}{f'}\frac{f'' \left(f - k f'\right) - f'\left(f' - f'- k f''\right)}{(f - k f)^2}\\ &= \frac{-k f'' f}{f'(f - kf)},\\ &= \frac{F_{K,L} F}{F_K F_L} \end{align*} $$
