Your question really depends on the values of x and y. And your expression doesn't have c. I assumed your function to be:
f(x,y)=\sqrt{a x^8+b x^4y^4+c y^8}
I only outline the method to obtain a power series here.
f(x,y)=c y^8\sqrt{1+\left(\frac{a x^8+b x^4y^4}{c y^8}\right)}
By
(1+x)^p=\sum _{k=0}^{\infty } \binom{p}{k}x^k
,we have:
f(x,y)=c y^8\sum _{k=0}^{\infty } \binom{\frac{1}{2}}{k}\left(\frac{a x^8+b x^4 y^4}{c y^8}\right)^k
There is another expansion method as well, in which you take out the power x instead of y. It depends on the values of x and y. Since the expansion of (1+x)^p is valid for |x| < 1 (|x| = 1 is more complicated).