about how to share one External serial EEprom using two microcontrollers?
You didn't show us a complete, minimal, compilable example, so my answer will only cover some important points.


Your indention is way off. It is important to use the correct indention, since it help structure your code and will let you quickly see misplaced brackets. The Arduino IDE even has an autoformat function for this.
You are configuring and checking always the whole PORT A, but you have only connected one of the pins. When you configure the whole port as input (without pullup), the other pins will be floating. Also, if both devices have the pin on input (no pullup), this pin will also be floating. The check cannot work this way. You should only look at the first bit in the port registers, since that is the pin, that you connected. You can set all other bits to zero when reading, by performing a bitwise AND operation: if(PINA &amp; (1&lt;&lt;PA0)). This will execute, when the first bit of PINA is 1. To set a single bit you can use the bitwise OR: DDRA |= 1&lt;&lt;PA0; will set the first bit of DDRA. To erase only that bit from the register, you can use DDRA &amp;= ~(1&lt;&lt;PA0);. The rest of the port will stay untoched with this.
Your current protocol is, that the first MCU will check for LOW, do it' transmission, then drive the pin actively HIGH as a signal for the second MCU to start transmitting. The second MCU will do it's I2C stuff and then drive the pin actively LOW. So now we have an active LOW and active HIGH on the same pin, which in principle is a short. A rather big amount of current will flow from one MCU to the other, most likely destroying at least one of the pins (meaning the output hardware in the MCU, that serves that pin). To avoid this, you can use a principle like that, that the I2C port uses: Open Drain logic. You can start with the first MCU's pin configured as OUTPUT LOW, while the other is INPUT PULLUP. LOW means the first MCU can work. If the first MCU want's the second MCU to do a transmission, it will release the line by setting the PIN to INPUT PULLUP. The second MCU now sees a HIGH level on it's pin and starts it's transmission. When it is finished, it will drive the line actively LOW for a defined duration (enough for the first MCU to react) and then to INPUT PULLUP again. The first MCU will then see a LOW level on it's pin as a signal, that the second MCU is finished. It then itself drives the line actively LOW. Then the play starts again. The important thing is, that a HIGH state is only done by a pullup, not actively driven.




That said, I still don't see the necessity to share an external EEPROM for this use case. You are overthinking the problem. You wrote, that you don't want to miss a frame, if the second MCU is still processing the last frame. There are better ways to do that (especially when you consider the limited lifetime of EEPROM).


You didn't tell us, why you are using 2 MCU's, so maybe your goal can be reached with only 1 MCU, which will eliminate the whole need for transmission.
You could use another UART interface via SoftwareSerial or it's sibling libraries to let the MCU's communicate directly. "Not missing a frame" will still be the case, when you are not disabling interrupts. The library will read the data into a buffer in interrupts, while you are processing the previous frame. Depending on how much data you send, you might want to make the buffer bigger, but that is a simple change in the libraries definitions.
If you want to use I2C to communicate between both MCU's, you have 2 ways of doing it. You can either go the same way as described above for UART, by just letting it fill a big enough buffer, or you can do this through the bus transmissions. The first MCU would be the master, that tries to send the data to the second MCU (slave), when there is new data available. The slave will receive the data normally. When the frame is finished receiving, the slave can leave the bus, until it is finished processing. So, when the master tries to send more data in this time, the transmission will fail at the start, since no slave with that address is on the bus (the slave MCU left the bus for this period of time). You can check the return value of Wire.endTransmission(), which will reflect the corresponding error code. If this error code is returned, the master should wait a bit and try to send this frame again, until it is received successfully. That also means, that the master needs a big enough buffer for storing new data, while waiting for the slave to be ready. Or another way: The master can request 1 byte from the slave, that indicates, if the slave is ready for new data. If not, the master will wait a bit and ask again. If yes, it will transmit the data.


Note: Either way the frequency of data reception and processing must be equal for continuous operation. The whole chain will be as fast as the slowest part. If you receive data faster, than it get's processed, the buffer will fill up fast, until you really have to drop frames. The prinicples above are more about the case, when the processing and receiving may misalign or vary a bit, but at the average staying at the same speed. And for this simple communication buffers, that all communication libraries use, with enough space for maybe 2 to 4 transmissions (depending on the variation) are way enough.