Remember my formula of E=(u/k)c
It seemed to me that the mathematicians here did not understand it.
So, I figure that a mathematical breakdown may help their understanding.

e=((c*m^2)/(u/m))*c

e = (c^2 m^3)/u

u = (c^2 m^3)/e, c m!=0

Reduce[(c^2 m^3)/u == E, {c, m, u}]

{{u == (c^2 m^3)/E, c m != 0}}

derivatives

((partial d)c(m,u))/((partial d)u) = e/(2 c m^3)

((partial d)c(m,u))/((partial d)m) = -(3 c)/(2 m)

((partial d)m(c,u))/((partial d)u) = e/(3 c^2 m^2)

((partial d)m(c,u))/((partial d)c) = -(2 m)/(3 c)

((partial d)u(c,m))/((partial d)m) = (3 c^2 m^2)/e

((partial d)u(c,m))/((partial d)c) = (2 c m^3)/e

((partial d)^2c(m,u))/((partial d)u^2) = -e/(4 c m^3 u)

((partial d)^2c(m,u))/((partial d)m (partial d)u) = -(3 e)/(4 c m^4)

((partial d)^2c(m,u))/((partial d)m^2) = (15 c)/(4 m^2)

((partial d)^2m(c,u))/((partial d)u^2) = -(2 e)/(9 c^2 m^2 u)

((partial d)^2m(c,u))/((partial d)c (partial d)u) = -(2 e)/(9 c^3 m^2)

((partial d)^2m(c,u))/((partial d)c^2) = (10 m)/(9 c^2)

((partial d)^2u(c,m))/((partial d)m^2) = (6 c^2 m)/e

((partial d)^2u(c,m))/((partial d)c (partial d)m) = (6 c m^2)/e

((partial d)^2u(c,m))/((partial d)c^2) = (2 m^3)/e

That is the way it is, and its very simple, once you understand it.
The question that is impossible for mathematics to answer is WHY is it this way.

Oops the mathematical bracket symbol does not work the way it should in this instance.
What can I do?