What is the transfer function $H_{xy}$ of this flow graph?
If you mark with $v[n]$ the signal between summing nodes for input $x[n]$ and output $y[n]$, then:

$v[n] = x[n] -8v[n-2] -4v[n-3]$

Re-arranging sides:

$v[n]+8v[n-2]+4v[n-3]=x[n] $

Using the Z-transform:

$V(z)[1+8z^{-2}+4z^{-3}]=X(z)$

Now let's calculate the transfer function between $v[n]$ and $x[n]$:


  $H_{x\to v}(z)=\dfrac{V(z)}{X(z)}=\dfrac{1}{1+8z^{-2}+4z^{-3}}$


Now moving to the relation between output $y[n]$ and $v[n]$:

$y[n]=v[n]+2v[n-1]$

Again calculating the Z-transform:

$Y(z)=V(z)[1+1+2z^{-1}]$

Now converting the transfer function $H_{vy}(z)$:


  $H_{v\to y}(z)=1+2z^{-1}$


Final step is to merge these two together. We know that in time domain it is a convolution: $h_{x\to y}[n]=h_{x\to v}[n]\star h_{v\to y}[n]$, thus in Z-transform domain it will be the multiplication, yielding the final result:


  $H_{x\to y}(z)=H_{x\to v}(z)H_{v\to y}(z)=\dfrac{1+2z^{-1}}{1+8z^{-2}+4z^{-3}}$