Its become easier to analyze the circuit when you forget the forward bias voltage for the diodes. ( Once you understand the trip points, you can then consider them later ).
Lets call the left Cell as V1 and the right one as V2.
So lets replace diodes by switch. Which is closed when forward biased and open when reverse.
Now if the voltage at the junction of R1 and R2 would be half of V2 ( V2 * (R1 + R2) / 2 ) when D2 is forward biased. Just keep that in mind.
Case 1: V2 < V1
No current flows through R1 since D2 is open. ( reverse biased ).
Current through R2 is entirely supplied by V1 so it could be
Ir2 = V1/R2
adjusting for assumptions:
Ir2 = (V1 - Vdf ) / R2; where Vdf is the forward drop across Diode.
Ir1 = 0; ( since D2 is reverse biased ).
Case 2: V2 > 2 times V1
Now V2 is powerful enough to create a voltage at the node of R2 and R1 so as to shut down D1.
Ir1 = Ir2 = V2 / (R1 + R2); ( since both are in series ).
Thus adjusting for assumptions,
Ir1 = Ir2 = (V2 - Vdf) / (R1 + R2);
Case 3: V2 > V1 but V2 < 2 times V1
This is the case where both the diodes are forward biased. For example as in your case, where V1 = 6V and V2 = 9 V.
Its quite easy to solve when you look at the current flowing through resistors R1 and R2 and the voltages across them.
Lets assume Id1 and Id2 are the currents flowing through the diodes. so we have,
Id2 is flowing thorugh R2, and Id1 + Id2 flowing through R1.
So we have two equations.
V1 = R2 * ( Id1 + Id2 );
V2 = V1 + ( Id2 * R2);
adjusting for forward drops;
V1 - Vdf = R2 ( Id1 + Id2);
V2 - Vdf = (V1 - Vdf ) + ( Id2 * R2 );
Since we have the values for R1 and R2 as R = 1K, putting them in the equation we have;
V1 - Vdf = R ( Id1 + Id2); -----( 1 )
V2 - Vdf = (V1 - Vdf ) + ( Id2 * R ); -----( 2 )
Solving for Id2 in equation ( 2 ) we have;
Id2 = ( V2 - V1 ) / R;
Substituting Id2 in the equation ( 1 ) we get,
V1 - Vdf = R ( Id1 + ((V2 - V1 ) / R ) );
that boils down to;
Id1 = ( 2V1 - V2 - Vdf ) / R;
Examples.
Solving for V2 = 9V and V1 = 6V , R = 1K and Vdf = 0.7Volts,
Id2 = (9 - 6 ) / 1000 A = 3mA;
Id1 = ( 2 * 6 - 9 - 0.7 ) / 1000 = (12 - 9 - 0.7 ) / 1000 = 2.3mA;
Solving for V2 = 9V and V1 = 7.5V , R = 1K and Vdf = 0.7Volts,
Id2 = (9 - 7.5 ) / 1000 A = 1.5mA;
Id1 = ( 2 * 7.5 - 9 - 0.7 ) / 1000 = (12 - 9 - 0.7 ) / 1000 = 5.3mA;