How to calculate Vx in this circuit?
Question 1: Assuming that D1 is off and D2 is on, we have the following circuit:


We can first calculate the current throught diode 2

\$ i_{D2} = \frac{9V-(-6V)}{22k \Omega + 43k \Omega} = 0.2308 mA \$

and then the voltage across diode 1

\$ V_{D1} = 9V-0.2308mA \cdot 22 k \Omega = 3.92 V \$

As a diode may only block negative voltages, the assumption of diode 1 being off and D2 being on is invalid.

Question 2: Assuming that both diodes are turned on yields the following circuit:


Calculating the current through diode 2

\$ i_{D2} = \frac{0V-(-6V)}{43 k \Omega} = 0.1395 mA \$

and then the current denoted as ix

\$ i_x = \frac{9V - 0V}{22 k \Omega} = 0.4091 mA \$

Using Kirchoffs Current Law (KCL) stating that all currents entering a node should be equal to the current leaving the node we get

\$ i_x = i_{D1} + i_{D2} \$

Which may be rearranged to

\$ i_{D1} = i_x - i_{D2} = 0.4091 mA - 0.1395 mA = 0.2696 mA \$