Here is a proof for a general HOD 1 function.

Exercise 3.3
This can actually be directly obtained by Euler Theorem. Or just go through the following proof.
HOD 1 implies f(tX) = tf(X) for any t>0......(1)
Differentiate w.r.t. x_{i}:
tf_{i}(tX)=tf_{i}(X), which then gives f_{i}(tX) = f_{i}(X) ......(2)
Differentiate w.r.t. t:
\sum_{i=1}^{n} f_{i}(tX)*x_{i} = f(X).......(3)
Combining the (2) & (3) equations, we have \sum_{i=1}^{n} MP_{i}(X)=f(X).