Is this a parallel circuit? Can I collapse it and make one equivalent resistor of \$2/3\Omega\$?
Also, will someone confirm whether or not I got the right answers for the unknown voltages and currents? \$I_o = 1.333\dots\text{A}\$, \$I_x=2.6666\dots\text{A}\$, \$V_o = 4\text{V}\$.
I found \$I_o\$ with current division: \$(1/3)\times(4) = 4/3 = 1.333\text{A}\$
and \$I_x = (2/3)\times 4 = 8/3 = 2.6666\text{A}\$
(Update: I see my mistake. I was multiplying by 4V instead of 6A when current dividing.)
Best Answer
Starting from one terminal of source, if current has more than one path to reach the other terminal, then those two paths are parallel. I see two such paths in your circuit.
You didn't say how you calculated \$I_x\$ and \$I_o\$. The answers you got are wrong. Try Current dividision.
EDIT: You used 4A instead of 6A in your calculations.
If you want to find the current by dividing voltage across resistance by resistance value, you have to find the voltage \$V_o\$ first. $$V_o = 6A\times (1\Omega || 2\Omega) = 4V$$ now,
$$I_x = V_o/1\Omega = 4A$$ $$I_o = V_o/2\Omega = 2A$$