I've got that G is isomorphic to C_6 x C_2 , and I have to prove that G has got only one subgroup (I'm going to call H to that subgroup) of order 3.
I've done the following:
I've taken an element from G, which is called (h,k), (h is an element from C_6 and k is an element from C_2). So the order of (h,k) is going to be: o((h,k)) = mcm (o(h),o(k)). So o(h) can be 1,2,3 or 6 and o(k) can be 1 or 2. But mcm (o(h),o(k)) = 3 if and only if o(h)=3 and o(k)=1. And I also know that as C_6 is cyclic, has got only one subgroup of order 3.
But now I don't know how to continue and which conclusion I have taken from everything I've done.
Am I maybe doing something wrong?
What you have done is correct. A subgroup of order $3$ is cyclic, generated by an element $(h,k)\in G$. Based on what you have done, we know that the order of $h$ is $3$ and the order of $k$ is $1$. Now, $\mathbf{Z}/6\mathbf{Z}$ has multiple elements of order $3$, but they all generate the same subgroup.
Alternatively, you could have noticed that using the Chinese remainder theorem, we have $G\simeq \mathbf{Z}/3\mathbf{Z}\times (\mathbf{Z}/2\mathbf{Z})^2.$