How to find the number of cyclic subgroups of order $3$ of $\mathbb{Z}_3 \times \mathbb{Z}_9$?And how to find total subgroups of this group? I have found the cyclic subgroups by first finding the order of $(3,1),(3,3),(1,3)$ and they are $\phi(3)+\phi(3).\phi(3)+\phi(3)=2+2.2+2=8$ and then divide it by $\phi(3)=2$.Then answer is 4.Am I right?

Your answer is correct but your approach seems flawed (or maybe requires clarification).
First of all the order of $(3,1) \in \mathbb{Z}_3 \times \mathbb{Z}_9$ which is same as $(0,1)$ is $9$ and NOT $\phi(3)$.  
To find the total number of subgroups of order $3$ of $\mathbb{Z}_3 \times \mathbb{Z}_9$ we need to find all the elements of order $3$ (because any group of order $3$ is cyclic, hence finding such elements is sufficient). 
Note that the order of $(a,b)$ is given by $\text{lcm}(|a|, |b|)$. Thus to obtain order $3$ elements we need to find $(a,b) \in \mathbb{Z}_3 \times \mathbb{Z}_9$ such that 
$$|a|=1 \text{ and } |b|=3 \quad \text{or} \quad |a|=3 \text{ and } |b|=1 \quad \text{or} \quad |a|=3 \text{ and } |b|=3.$$
Now 


*

*the number of elements $(a,b)$ such that $|a|=1 \text{ and } |b|=3$ are $1 \cdot \phi(3)=2.$

*the number of elements $(a,b)$ such that $|a|=3 \text{ and } |b|=1$ are $\phi(3) \cdot 1=2.$

*the number of elements $(a,b)$ such that $|a|=3 \text{ and } |b|=3$ are $\phi(3) \cdot \phi(3)=4.$


Thus there are a total of $8$ elements of order $3$. However the number of subgroups of order $3$ will be $4$ because e.g. both $(0,3)$ and $(0,6)$ will generate the same subgroup.
