What is the transfer function of this block diagram I and my friend have different answer for this block diagram



Mine:
$$y[n] = -\frac23x[n] + x[n-1] -\frac12[n-2]$$

Hence

$$H(z) = \frac{1}{-\frac23 + z^{-1} -\frac12z^{-2}}$$

My friend:
$$q[n]=x[n]-\frac12q[n-1]$$
$$y[n] = -\frac23x[n]+q[n-1]$$
$$Y(z) = \frac23X(z)+z^{-1}Q(z)=\frac23X(z)+z^{-1}\frac{X(z)}{1+\frac12z^{-1}}$$

Hence

$$H(z)=\frac{Y(z)}{X(z)}=\frac23+\frac{z^{-1}}{1+\frac12z^{-1}} = \frac{4z^{-1}+2}{3+\frac32z^{-1}}$$

I don't know how can he come to that result but I'm not sure about mine either
As pointed out in the other answers, your friend's approach is correct. However, his end result for $H(z)$ isn't.

With $q[n]$ the input signal of the delay element, you get $q[n-1]$ at its output, and the given difference equation for $q[n]$ follows easily. The output $y[n]$ can then be written in terms of $x[n]$ and $q[n-1]$. In the $\mathcal{Z}$-transform domain you get algebraic equations in terms of $X(z)$, $Q(z)$, and $Y(z)$, from which you can eliminate $Q(z)$, leaving you with an expression for $H(z)=Y(z)/X(z)$:

$$H(z)=\frac{Y(z)}{X(z)}=-\frac23+\frac{z^{-1}}{1+\frac12 z^{-1}}=-\frac23\frac{1-z^{-1}}{1+\frac12 z^{-1}}\tag{1}$$

So it was basically just a sign error that messed up the end result.