Can anyone explain me how the voltage divider network works?
Try to use superposition principle. 

First set 3.3V to 0V and solve for \$V_O'\$







simulate this circuit &ndash; Schematic created using CircuitLab

$$V_O' = 5V*\frac{25k||25k}{25k||25k + 200k} = 5V*\frac{12.5k}{12.5k + 200k} = 0.2941V  $$
Next turn on 3.3V and turn off 5V source (set to 0V)







simulate this circuit

$$V_O''= 3.3V * \frac{25k||200k}{25k+25k||200k} =  3.3V\frac{22.22k}{22.22k + 25k} = 1.552V$$ 

and finally we have the answer $$V_O=V_O'+V_O'' = 0.294 + 1.552V = 1.846V$$

Or try to do nodal analysis. 

$$\frac{3.3V - V_O}{25k\Omega} + \frac{5V - V_O}{200k\Omega}=\frac{V_O}{25k\Omega} $$  And solve for \$V_O\$