What currents are entering/exiting this end of the supernode?
First thing, you should predefine the direction of all currents. If the final value of the current is negative, it means that you have to flip the originally defined current direction.



Considering KCL of each voltage loop

$$14V = i_1\cdot 4\Omega  + i_3\cdot 3\Omega $$

$$6V = i_2\cdot 1.5\Omega  - i_3\cdot 3\Omega $$

$$ i_1 = i_2 + i_3$$

Rearranging the equations you obtain:

$$i_1 = 3.6A$$

$$i_3 = -0.13A$$

$$i_1 = 3.73A$$

It means that at the node of interest, the current I_3 is flowing upwards and the current I2 coming from the parallel connection of the right most resistors are entering the node from the right side.