Despite what most textbooks claim, superposition of dependent sources is valid if done correctly.
There are three sources in this circuit so there will be three terms in the superposition.
For the first term, the two current sources are zeroed (opened) so \$V_x\$ is given by voltage division:
\$V_x = 10V \cdot \dfrac{4}{4 + 20} = \dfrac{5}{3}V\$
For the second term, the voltage source is zeroed (shorted), so the two resistors are now in parallel, and the 2A source is activated. Thus:
\$V_x = 2A \cdot 4\Omega || 20 \Omega = \dfrac{20}{3}V\$
Since the third, dependent source is in parallel with the 2A source, the last term has the same form:
\$V_x = 0.1 V_x \cdot 4\Omega || 20 \Omega = \dfrac{1}{3}V_x\$
Now, it's crucial at this point to not try and solve the previous equation (you'll only get \$V_x = 0\$ if you do.)
Rather, proceed with the superposition sum and then solve.
\$V_x = \dfrac{5}{3}V + \dfrac{20}{3}V + \dfrac{1}{3}V_x\$
Grouping terms:
\$V_x (1 - \frac{1}{3}) = \dfrac{25}{3}V\$
Solving:
\$V_x = 12.5V\$
The step where you canceled out (2j + 1) / 2j does not look right.
Subsequent values are going to change. In any case:
1 / (1 + 2j) does not equal to (1 + 1/2 j).
One standard way of getting rid of a complex value in the denominator is to multiply top and bottom with the complex conjugate of the denominator.
When everything are fixed, convert final answer to polar form and there you have the amplitude and phase.
Best Answer
a) Your start equation is missing a factor 6, which is the value of the resistor.
The frequency response would then be:
$$ G(S)=\frac{Y(S)}{U(S)}=\frac{3.(S+2)}{2.S+1} $$ b)When the the frequency is zero the cap is an open circuit and the resulting gain should be 6, which is the value of the the resisor is parallel with the source.
When the frequency approaches infinity, the cap is a short circuit and the gain should be the parallel combination of the two resistor. That would be 12/8, you will reach the same value after calculating the limit with S-->infinity of the equation above.