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Let $\langle xN,yN \rangle$ be a subgroup of $G/N$.
Then the preimage of $\langle xN,yN \rangle$ under natural map is $\langle x,y\rangle$.
I am trying not to use correspondence theorem.

Let $v:G\rightarrow G/N$ be the natural map.
Let $K$ be the preimage of $\langle xN,yN \rangle$ under $v$.
Clearly $x,y\in K$ since $v(x)=xN$ and $v(y)=yN$.
Thus $\langle x,y\rangle\subseteq K$.

For reverse inclusion, let $k\in K$.
Then $v(k)=kN\in\langle xN,yN \rangle$.
So $kN=zN$ where $z\in\langle x,y\rangle$.
Then $k=zn$ for some $n\in N$.
So for this part the proof is not complete.

I think the part missing is the fact that $N\subseteq \langle x,y\rangle$. But I can only prove that $N\subseteq K$.

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This isn't true. For instance, if $x=y=1$, then $\langle xN,yN\rangle$ is the trivial subgroup and $K=N$. But $\langle x,y\rangle$ is not equal to $N$ unless $N$ is trivial.

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    So under what condition will this be true? As I am trying to find the preimage of $\langle sZ(D_{16}),r^2Z(D_{16})\rangle$ which is a subgroup of $D_{16}/Z(D_{16})$ where $D_{16}=\langle r,s \;|\; r^8=s^2=1,rs=sr^{-1} \rangle$ – 2017-01-31
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    It should be $(z,r^2)Z(D_{16})$. The condition your after is if the subgroup contains $N$. – 2017-01-31
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    It will be true iff $N\subseteq\langle x,y\rangle$...I doubt you can get a more concrete condition than that. – 2017-01-31