Answer
First Case:
Assume $\ f^-1$(C U D) is true
by def of range f(y) = x , x$\in$(C U D) **( x $\in$ C or x $\in$ D by def of union)
$\ f^-1$(y) = x , x$\in$ C implies x$\in$$\ f^-1$(C)
or
$\ f^-1$(y) = x , x$\in$ D implies x$\in$$\ f^-1$(D)
by def of union x $\in$ $\ f^-1$(C) U $\ f^-1$(D)
Second Case :
Assume $\ f^-1$(C) U x$\in$$\ f^-1$(D) is true
hence, x $\in$ $\ f^-1$(C) U $\ f^-1$(D)
x$\in$ $\ f^-1$(C) $OR $ x$\in$$\ f^-1$(D) **By definition of union
which implies $\ f^-1$(y) = x , x $\in$ (C U D)
which also means x $\in$ $\ f^-1$(C U D)
In conclusion, we got from :
Case 1: $\ f^-1$(C U D) $\subseteq$ $\ f^-1$(C) U $\ f^-1$(D)
Case 2: $\ f^-1$(C) U $\ f^-1$(D) $\subseteq$ $\ f^-1$(C U D)
hence, $\ f^-1$(C U D) $=$ $\ f^-1$(C) U $\ f^-1$(D)
Work Step by Step
First Case:
Assume $\ f^-1$(C U D) is true
by def of range f(y) = x , x$\in$(C U D) **( x $\in$ C or x $\in$ D by def of union)
$\ f^-1$(y) = x , x$\in$ C implies x$\in$$\ f^-1$(C)
or
$\ f^-1$(y) = x , x$\in$ D implies x$\in$$\ f^-1$(D)
by def of union x $\in$ $\ f^-1$(C) U $\ f^-1$(D)
Second Case :
Assume $\ f^-1$(C) U x$\in$$\ f^-1$(D) is true
hence, x $\in$ $\ f^-1$(C) U $\ f^-1$(D)
x$\in$ $\ f^-1$(C) $OR $ x$\in$$\ f^-1$(D) **By definition of union
which implies $\ f^-1$(y) = x , x $\in$ (C U D)
which also means x $\in$ $\ f^-1$(C U D)
In conclusion, we got from :
Case 1: $\ f^-1$(C U D) $\subseteq$ $\ f^-1$(C) U $\ f^-1$(D)
Case 2: $\ f^-1$(C) U $\ f^-1$(D) $\subseteq$ $\ f^-1$(C U D)
hence, $\ f^-1$(C U D) $=$ $\ f^-1$(C) U $\ f^-1$(D)