Why can’t i use a supermesh analysis on this circuit?
Just re-draw the schematic (note that I chose a specific node for "ground" as the (-) terminal of the \$+12\:\text{V}\$ voltage source.)







simulate this circuit &ndash; Schematic created using CircuitLab

The whole thing is a lot less complicated, now.

To see how I started to produce the above schematic, see my modifications to your schematic below:



You can use KCL to compute \$V_\text{X}\$ as (currents flowing out of the node on the left side of equals, currents flowing into the node on the right side of equals):

$$\frac{V_\text{X}}{R_1}+I_1 + 5\:\Omega^{-1}\left(12\:\text{V}-V_\text{X}\right)=\frac{12\:\text{V}}{R_1}$$

Solve that for \$V_\text{X}\$.

Now use KCL to compute \$V_\text{Y}\$ as (currents flowing out of the node on the left, currents flowing into the node on the right):

$$\frac{V_\text{Y}}{R_2}+\frac{V_\text{Y}}{R_2}=\frac{0\:\text{V}}{R_2}+\frac{12\:\text{V}}{R_3}+5\:\Omega^{-1}\left(12\:\text{V}-V_\text{X}\right)$$

Solve that for \$V_\text{Y}\$.

Now simply compute \$I_0=\frac{12\:\text{V}-V_\text{Y}}{R_3}\$.

You will find the same answer as you gave.