How to solve this circuit problem?

Notice first that Vo = 2Ω*io. This has to be true because all of these elements are in parallel. Their top ends are connected and their bottom ends are connected so they all have the same voltage across them.

Using KCL we can write:

6 = io+io/4 + (2*io)/8 = 6*io/4 = 3*io/2 => 12 = 3*io => io = 4

Vo = 2*io = 8V

Let;

I1 = 6A source

I2 = ie/4 amps

R1 = 2 Ohms

R2 = 8 Ohms

Vo = node voltage at junction at top of diagram

Ground reference at bottom

Solve for Vo;

Vo*(1/R1 + 1/R2) - I1 + I2 = 0

I2 = ie/4

Vo*(1/R1 + 1/R2) - I1 + ie/4 = 0

ie = Vo/R1

Vo*(1/R1 + 1/R2) - I1 + Vo/(4*R1) = 0

collect terms;

Vo*(1/R1 + 1/R2 + 1/(4*R1 ) - I1 = 0

Vo*(1/R1 + 1/(4*R1) + 1/R2 ) - I1 = 0

V0*[5/(4*R1) + 1/R2] - I1 = 0

V0*[5/(4*R1) + 1/R2] = I1

V0 = I1/[5/(4*R1) + 1/R2]

Vo = 6/(5/8+1/8)

Vo = 8 V .................. Answer

ie = 8/2

ie = 4 A .................. Answer

It seems to me that the answer is printed right there in the question.