thevenin dependent source finding Rth
This is one of those pesky cases where the standard procedure for finding the open circuit voltage or the short circuit current leads to "funny" results.

Another way to compute the Thevenin equivalent resistance of a network N is the following:


(1) Disable all independent voltage/current sources inside N, i.e. current sources replaced by open circuits, voltage sources by short circuits. Dependent sources remain untouched.
(2) Apply an (ideal) voltage \$V_0\$ (current \$I_0\$) source across the terminals of N.
(3) Compute the current \$I_0\$ entering (the voltage \$V_0\$ across) the terminals of N
(4) Compute the equivalent resistance as \$ R_{th} = \dfrac{V_0}{I_0} \$


In our case, applying a voltage source \$V_0\$ to the output makes \$V_x=V_0\$, thus it is easy to see that the current entering the network is:

\$
I_0 = \dfrac{V_0 - \mu V_0}{R} = (1-\mu) \dfrac{V_0}{R}
\$

Hence:

\$
R_{th} = \dfrac{V_0}{I_0} = \dfrac{R}{1 -\mu} = \dfrac{8 \Omega}{1 - 0.5} = 16\Omega
\$