The bezier surface is the function of parameter like x = f(u,v).
I have the function of biquadratic bezier surface which contains 9 control points x1 to x9.
$$((1-u)^2)((1-v)^2)x_1+2u(1-u)((1-v)^2)x_2+(u^2)((1-v)^2)x_3+2v((1-u)^2)(1-v)x_4+4uv(1-u)(1-v)x_5+2(u^2)v(1-v)x_6+(v^2)(1-u)^2x_7+2(v^2)u(1-u)x_8+(u^2)(v^2)x_9==x,$$
$$((1-u)^2)*((1-v)^2)*y1+2*u*(1-u)* ((1-v)^2)*y2+(u^2)*((1-v)^2)*y3+2*v*((1-u)^2)*(1-v)*y4+4*u*v*(1-u)*(1-v)*y5+2*(u^2)*v*(1-v)*y6+(v^2)*(1-u)^2*y7+2*(v^2)*u*(1-u)*y8+(u^2)*(v^2)*y9==y,$$
$$((1-u)^2)*((1-v)^2)*z1+2*u*(1-u)* ((1-v)^2)*z2+(u^2)*((1-v)^2)*z3+2*v*((1-u)^2)*(1-v)*z4+4*u*v*(1-u)*(1-v)*z5+2*(u^2)*v*(1-v)*z6+(v^2)*(1-u)^2*z7+2*(v^2)*u*(1-u)*z8+(u^2)*(v^2)*z9==z$$
this is like
$$x = f1(u,v)$$
$$y = f2(u,v)$$
$$z = f3(u,v)$$
How can I eliminate parameters u and v to get the function of x,y,z?
like $$f(x,y,z)=a*x^2+b*y^2+c*z^2+d*xy+e*xz+f*yz+g*x+h*y+i*z+k$$
Look up terms like "elimination" and "resultant".
The answer will be a polynomial of degree 8, and the algebra will be pretty complicated. I'd recommend using a computer algebra system like Mathematica or Maple.
This paper (and its references) might be a good place to start. Or, try this.