Since the current through R2 is zero, Vo = Vx. You can calculate Vo (and Vx) just using the known expression of a voltage divider:
$$
V_o = V_x = \frac{\frac{1}{sC}}{\frac{1}{sC} + R_1} V_i= \frac{1}{R_1Cs + 1} V_i.
$$
it appears to me that the current generated by 35 V and 2vx will
collide each other
It may be that you are assuming that a voltage source, whether independent or controlled, must source current, i.e., supply power to the circuit.
But, at least in ideal circuit theory, there's nothing "wrong" with a voltage source sinking current, i.e., receiving power from the circuit.
For a real world example, consider that, when a battery is being charged, the current is in the opposite direction than when the battery is being discharged.
I would like to know how the current flows across 5 Ω resistor.
If you're planning to be an EE, don't write or say things like "current across"; current is through, voltage is across.
Now, this circuit is very easy to solve. There are two unknowns so you need two independent equations.
For the 1st, write a KVL equation clockwise 'round the loop:
$$35V = v_x + 2v_x - v_o \rightarrow 3v_x = 35V + v_o$$
Now, you need one more independent equation. Can you find one?
Best Answer
The sign of the source is drawn incorrectly or the polarity of how \$V_x\$ is defined should be reversed.
And most likely the equation is correct.
$$ Vx + 3Vx + 6i - 20 = 0 $$
together with the relation $$ V_x=3\cdot i$$
yields
$$ 18i - 20 = 0 $$
If it were $$ Vx - 3Vx + 6i - 20 = 0 $$ it would yield $$ -20=0$$