Boolean expression to logic circuit

A two-level implementation with NOR gates requires that the function is simplified into product-of-sums (POS) form.

Your function \$F=AB+C' \$ is not in that form. Convert it into POS form, invert it twice, and use DeMorgan's Law to obtain a function that is realizable using only NOR gates.

Edit

The first part is to convert the your function into POS form. This can be done elegantly by using the AND-distributive law

$$F=(A+C')(B+C') \tag1$$

Inverting this once gives

$$F'=[(A+C')(B+C')]'=(A+C')'+(B+C')' \tag2$$

All you have to do is invert the expression one last time, and you can realise the function with only NOR gates.

Edit 2

From the comments

i have already gotten the function down to the last step in the second last image i have been told that much is true, I think my actual software schematic may have been unreadable as when i test it against the truth table it all checks out

I think you have worked on this long enough. I will show you my solution.

Inverting \$F' \$ from the first edit leaves us with the final expression. $$F''=F=[(A+C')'+(B+C')']' \tag3$$

So the circuit can be realized with 4 NOR gates like this.

Note that we here use that a NOR gate with duplicate inputs acts as an inverter. However, this is necessary as OP only can use NOR gates.