This is the Implicit Function Theorem taken to the 2nd derivative.
So start with the function: #f(L, K(L)) = "const"#
The total derivative has 2 partials in it:
#(df)/(dL) = f_L + f_K \ K' = 0 #
#implies K' = - (f_L)/(f_K)#
Going again:
#(d^2f)/(dL^2) = d/(dL) (color(red)( f_L) + color(blue)(f_K \ K') ) #
#=color(red)( f_(LL) + f_(LK) \K') + color(blue)( (f_(KL) + f_(KK) \
K' )\ K' + f_K \K'') color(green)(= 0)#
#K'' =- (( K' (f_(LK) + f_(KL) )) + (K')^2 f_(KK) + f_(LL) ) / f_K#
#K'' =- (( - (f_L)/(f_K) (f_(LK) + f_(KL) )) + (- (f_L)/(f_K))^2 f_(KK) + f_(LL) ) / f_K#
#K'' =- (( f_(LL)f_K - f_L (f_(LK) + f_(KL) )) + (f_L^2 f_(KK))/(f_K) ) / (f_K)^2#
Mixed partial are same so:
#K'' =- ( f_(LL)f_K - 2 f_L f_(LK) + (f_L^2 f_(KK))/(f_K) ) / (f_K)^2#
