TL;DR

Special 0 rule

Rule:
0×0=0
Think like this,×n is time to type a number
example:
1×2=2 is "type number 1 2 times then sum"
So 0×0=0 is "type number 0 0 time then sum" the result is 0 because it's 0 time so you're not type to the input bar

0÷0=1:
Its like "how much time to type 0 so the result is 0" the result is 1 because 0 is already 0,if it's 0 then you're not type the number to input bar

0²÷0:
x²÷x,the ÷x neutrilize the ² so the result is x

Example:
x=0
Standard:
lim r->0,μ->0 μ²÷r,result is 0÷0=indeterminate->Hôpital thing->0
F(x)=2x÷x=(2×0)÷0=0÷0=indeterminate
G(x)=x²÷x:
0²÷0=indeterminate
Non standard(zero exclusion):
0²÷0,÷0 neutrilize the ²(x²÷x=x) so the result is 0
F(x)=2x÷x=2×0=0÷0=1
G(x)=x²÷x:
÷x neutrilize the ² like in 2²÷2=2,so 0²÷0=0

We use it a lot! So trying to redefine what it is is a very hard uphill struggle.

So I will just answer using the standard definition of numbers (and 0).

0 is already special. It is the additive identity in the Complex, Real, Integer, Natural, etc. number systems.

We use it a lot! So trying to redefine what it is is a very hard uphill struggle.

So I will just answer using the standard definition of numbers (and 0).

This is correct.

This is incorrect. Division by 0 is undefined.

Again, division by 0 is undefined.

The real numbers are a field and fields have an additive identity ##0## such that ##a+0=a## for all ##a## as well as a multiplicative identity ##1\ne 0## such that ##a \cdot 1 = a## for all ##a##. These lead to an additive inverse ##-a## such that ##a+(-a)=0## for all ##a## and a multiplicative inverse ##a^{-1}## such that ##a \cdot a^{-1}=1## for all ##a\ne 0##. Division is just shorthand for multiplication by the inverse ##a/b=a \cdot b^{-1}##.

From these you can prove that ##0 \cdot a = 0## for all ##a##:
$$0 \cdot a = (0+0) \cdot a$$$$0 \cdot a = 0 \cdot a + 0 \cdot a$$$$0 \cdot a + (- 0 \cdot a) = 0 \cdot a + 0 \cdot a + (- 0 \cdot a)$$$$ 0 = 0 \cdot a$$

Then the issue with assuming that ##0## has a multiplicative inverse is immediately apparent. Assume for proof by contradiction that ##0^{-1}## exists. By the multiplicative inverse definition ##0 \cdot 0^{-1}=1## but by the previous proof ##0 \cdot 0^{-1} = 0##. So ##0=1## which is a contradiction. Therefore ##0^{-1}## does not exist for the real numbers (or any field), so no expression of the form ##a\cdot 0^{-1}=b## is valid.

There are other number systems (not real numbers) where ##0^{-1}## does exist, but those numbers are not fields. So they lose a lot of the desirable proofs and properties that rely on being a field.

More briefly in your notation, if ##0/0=1## then $$1=0/0=(0\cdot 0)/0=0\cdot (0/0)=0\cdot 1=0$$ which is a contradiction.

You can make such an exception, but not with the real numbers.

The real numbers are a field and fields have an additive identity ##0## such that ##a+0=a## for all ##a## as well as a multiplicative identity ##1\ne 0## such that ##a \cdot 1 = a## for all ##a##. These lead to an additive inverse ##-a## such that ##a+(-a)=0## for all ##a## and a multiplicative inverse ##a^{-1}## such that ##a \cdot a^{-1}=1## for all ##a\ne 0##. Division is just shorthand for multiplication by the inverse ##a/b=a \cdot b^{-1}##.

From these you can prove that ##0 \cdot a = 0## for all ##a##:
$$0 \cdot a = (0+0) \cdot a$$$$0 \cdot a = 0 \cdot a + 0 \cdot a$$$$0 \cdot a + (- 0 \cdot a) = 0 \cdot a + 0 \cdot a + (- 0 \cdot a)$$$$ 0 = 0 \cdot a$$

Then the issue with assuming that ##0## has a multiplicative inverse is immediately apparent. Assume for proof by contradiction that ##0^{-1}## exists. By the multiplicative inverse definition ##0 \cdot 0^{-1}=1## but by the previous proof ##0 \cdot 0^{-1} = 0##. So ##0=1## which is a contradiction. Therefore ##0^{-1}## does not exist for the real numbers (or any field).

There are other number systems (not real numbers) where ##0^{-1}## does exist, but those numbers are not fields. So they lose a lot of the desirable proofs and properties that rely on being a field.

I showed why the custom rule you consider does not work for a field. So you can have either your custom rule or a field, not both.

I understand that.

I showed why the custom rule you consider does not work for a field. So you can have either your custom rule or a field, not both.

But what if it's PNEMDAS so "neutralize" before ()

Neutralize definition:
Divide number with exponent so

(x³)÷(x²):
÷x neutralize the ² so the result is x,if it's (x³) then ÷(x²)

example:
(x³)÷(x²)
÷(x²) neutralize the ³ so the result is x

(2³)÷(2²)
8÷4=2

So 0²÷0:
÷0 neutralize the ² so the result is 0,0×0÷0 then ÷0 neutralize the ×0 so the result is 0

There is nothing wrong with investigating your own ideas as long as you don't go too far down a "rabbit hole" for no reason. But this forum is not a good place for that because we have rules which forbid indulging in unpublished personal theories.
I will leave further discussion to others.

1×1=1,there's exception in standard math so there's exception for 0 too, same as custom rule,the exception is for certain number not for every number,if u force 2×2=2 then it's incorrect

Then all of the previous books and papers based on the standard axioms of fields would have to be discarded. And since that includes the real numbers that would be a lot. It would require the biggest book burning in history by far.

No. The field axioms are well established. Trashing them is a non-starter. You simply need to accept that your concept is incompatible with the real numbers. If you wish to use your custom exception then you cannot use the real numbers.