Inverse of \begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}
is given by
$\begin{bmatrix}
1 & -x & xz-y \\
0 & 1 & -z \\
0 & 0 & 1
\end{bmatrix}\tag{1}$
Denote the center of this group by $Z\big(G\big)=\left\{\, \begin{bmatrix}
1 & 0 & y \\
0 &1 &0 \\
0 & 0 & 1
\end{bmatrix} \quad \middle| \quad \text{ for any } y\in \mathbb{R} \, \right \}.$
Now show that $Z(G)$ is normal in $G$.
Let $A\in G, B\in Z(G)$ be any arbitrary element. Now show that $ABA^{-1}\in Z(G)$
$G$ is also known as Heisenberg Group(in our case over $\mathbb{R}$
Question by OP in comments: How to find $Z(G)$
Let $M=\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}\in Z(G)$ and $A=\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}$ be an arbitrary element in $G$. By definition we have $AM=MA$
$$\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix} =\begin{bmatrix}
1 & x & y \\
0 &1 &z \\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
1 & a & b \\
0 &1 &c \\
0 & 0 & 1
\end{bmatrix}.$$
Take the product.
$$\begin{bmatrix}
1 & x+a & y+az+b \\
0 &1 &z+c \\
0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
1 & a+x & b+cx+y \\
0 &1 &c+z \\
0 & 0 & 1
\end{bmatrix}.$$
Since $A$ was arbitrarily chosen. As a particular case,
Take $a=0 , c=1$ you get $x=0$
Take $a=1,c=0$ you get $z=0$