what is 4^6*x^8y^-2/ 4^7*x^4y^2

what is 4^6*x^8y^-2/ 4^7*x^4y^2

So we have

[(4^6)*(x^8)*(y^-2)] / [(4^7) * (x^4) 8 (y^2)]

So, by the property of exponents that says that  (a^n) /  (a^m)   =  a^(n-m)  We can write

[4^(6-7)] * [x^ (8-4)] * [y^(-2-2)]  =

[4^(-1)] * [x^4] * [y^(-4)] =

[x^4] / [4 * y^(4)]

$$\frac{4^6\times x^8 \times y^{-2}}{4^7}\times x^4y^2\\\\
=\frac{x^8 \times y^{-2}}{4}\times x^4y^2\\\\
=\frac{x^{12}}{4}$$

I have done it exactly as you have written it.

CPhill has done it as you quite likely intended. BUT if that is what you wanted then the stuff after the divide needed to be in brackets.