If we assume initially that the object is a cube then the surface area=6a^2, where a is the side of the cube.
So 6a^2=96x^4y^2z^8 and a^2=16x^4y^2z^8. Therefore a=4x^2yz^4.
The volume of the cube is a^3=64x^6y^3z^12.
If the object is a rectangular prism (cuboid) with sides a, b and c, and a=mx^2yz^4, b=nx^2yz^4 and c=px^2yz^4, such that 2(mn+mp+np)=96, mn+mp+np=48, and m=(48-np)/(n+p). It follows that 48-np>0 so np<48. If, for example, n=p=6, then m=12/12=1, and the volume is abc=x^2yz^4*6x^2yz^8*6x^2yz^4=36x^6y^3z^12. And m, n and p do not have to be integers. So the volume is variable and can be written: V(n,p)=np(48-np)x^6y^3z^12/(n+p) where np<48. V is maximum when n=p=4.
If the object is a rectangular pyramid with base sides a and b and perpendicular height c. V=(1/3)abc. The surface area, S, is ab+as1+bs2 where s1 and s2 are the different slant heights of opposite triangular faces. s1=√(c^2+b^2/4) and s2=√(c^2+a^2/4). So S=ab+a√(c^2+b^2/4)+b√(c^2+a^2/4)=96x^4y^2z^8. If a, b and c are defined as for the rectangular prism, S=(mn+m√(p^2+n^2/4)+n√(p^2+m^2/4))x^4y^2z^8=96x^4y^2z^8, so mn+m√(p^2+n^2/4)+n√(p^2+m^2/4)=96.
If m=n=p, S=m^2+m^2√5=96, m^2=96/(√5+1)=96(√5-1)/4=24(√5-1), and m=√(24(√5-1))=5.4466 approx. This makes a=b=c=√(24(√5-1)))x^2yz^4 and V=(1/3)abc=53.8591x^6y^3z^12 (approx).
If the shape is a sphere, 4πr^2=surface area=96x^4y^2z^8 and r^2=24x^4y^2z^8/π, so r=2√(6/π)x^2yz^4 and the volume is 4πr^3/3=(4π/3)48√(6/π)x^6y^3z^12/π=64√(6/π)x^6y^3z^12.
It's clear that the shape determines the complexity of the solution.
The question doesn't state the shape of the object, so the question is open-ended.