Raspberry PI 3 Model B Backup Battery
Your attempt doesn't even begin to meet the requirements of each component in the system, which are:-


Raspberry Pi model B: Regulated 5V DC supply at up to 1.7 Amps.
Lithium-ion cell: maximum voltage 4.20V. Maximum charge current ? (depends on battery capacity and 'C' rating). Minimum voltage under load 3.0V.     
Solar panel: efficient transfer of energy to battery and/or load (unless you don't mind using an oversized panel).


To run the Pi off a single Li-ion cell you absolutely need a voltage booster. This booster must supply up to 1.7A at 5V, so it may draw up to ~3.3A from the battery (assuming 85% converter efficiency).

To run the Pi off the solar panel without draining the battery, the panel must supply up to 10W to the converter. In practice this means you probably want at least a 20W panel. 

A voltage booster can only increase (not decrease) voltage, so if the panel puts out more than 5v you need a buck converter to drop it down to 5V or less. To charge the battery you must drop the panel voltage to 4.2V and apply current limiting. 

But solar panels automatically limit current according to light intensity, so your circuit also has to work properly when the panel cannot supply the set charge current. When this happens you need to throttle the charging current back to prevent the panel voltage from dropping below its maximum power point. Converters that do this are called MPPT (Maximum Power Point Tracking).   

To avoid damaging the battery you need some way to shut it off before becoming completely drained, and probably some warning so the Pi can be shut down while it still has power. When switched off your circuit must draw negligible current from the battery.

Finally, you want sufficient battery capacity to run the Pi for a reasonable period of time. The battery capacity you chose will effect the maximum charging current and how it can be applied. 

If the battery can take more current than the Pi draws then you could just hook the charger and booster straight to it without steering diodes. Then the battery will 'float' at up to 4.2V, while any excess current is absorbed by the booster (if it needs it). If the battery can only handle low charge current then it must be isolated from the output while being charged, and you need a bypass to route excess panel current to the booster. Your attempt embodies the basic concept of this configuration, but does nothing to address the other requirements. 

Bottom line is you cannot create a practical solar-powered UPS from two diodes and a resistor!