I think, question is split up into 2 parts. Part of this question is given in this link :- https://gateoverflow.in/94630/gate1988-13ia
So, here, it is given that G is the group of non zero real numbers under multiplication.
As we know that , In homomorphism ,
if mapping f : (G,*) ---> (H,*) for groups G and H
then f(a*b) = f(a)*f(b) for all a,b ∈ G
I assumed that mapping is :- f : (G,*) ---> (G,*)
So, let , y and z are the 2 elements of G such that y = p/q and z = r/s and p,q,r,s all are non-zero integers because G is a set of non-zero real numbers.
Since , here mapping is defined as f(x) = x^{3}
So, f(a*b) = f(a)*f(b)
f(y*z) = f(y)*f(z)
f((p/q)*(r/s)) = f(p/q)*f(r/s)
(pr/qs)^{3} = (p/q)^{3} * (r/s)^{3}
(pr/qs)^{3} = (pr/qs)^{3}
So, we can say that given mapping is Homomorphism.
According to definition of kernel ,
ker f = {g ∈ G : f(g) = e_{H}}
Since , identity element for multiplication operation on non-zero real numbers is 1
so , e_{H }= 1
and f(1) = 1^{3} = 1
So, g = 1
So, ker f = 1
https://en.wikipedia.org/wiki/Homomorphism
https://en.wikipedia.org/wiki/Kernel_(algebra)