Yes, 3.3k is the load. Write the equations for the voltage on the load, one for the -5V input and one for the +3V input (use Thévenin to find the equivalent V,R of everything but the load). These equations are equal to your target load voltages 0V, 3.3V respectively. Now you have two equations with two variables, R1, R2 which you can solve.
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Just remove the voltage source from the circuit and instead label the bottom left corner as 0V, the top left corner as -5V, and the top right corner as 0V. Then it's a simple voltage divider and algebra. Reduce it down to a simple function of R1 and R2...(hint: since the top and bottom right corners are 0V, you can knock the 3.3k resistor out and you just have a straight line...5V -> R2 -> 0V -> R1+470 -> -5V).
Then change the top left corner to 3V, the top right corner to 3.3V, and again reduce it down to a simple function of R1 and R2.
Finally, solve one of the two equations for either R1 or R2, plug it into the other one, solve for the variable, plug it back into either equation, and solve for the other variable.