First, we can use the chain rule for Jacobians to see how one of the terms in parentheses transforms: $$\frac{\partial (x, y)}{\partial (u, v)} = \frac{\partial (x, y)}{\partial (u', v')} \frac{\partial (u', v')}{\partial (u, v)}$$ A similar story holds for the other two factors. Combining them, we conclude that $$\sqrt{ \left( \frac{\partial (x,y)}{\partial (u,v)} \right)^2 + \left( \frac{\partial (y,z)}{\partial (u,v)} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u,v)} \right)^2 } =$$ $$\sqrt{ \left( \frac{\partial (x,y)}{\partial (u',v')} \frac{\partial (u', v')}{\partial (u, v)} \right)^2 + \left( \frac{\partial (y,z)}{\partial (u',v')} \frac{\partial (u', v')}{\partial (u, v)} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u',v')} \frac{\partial (u', v')}{\partial (u, v)} \right)^2 } =$$ $$\frac{\partial (u', v')}{\partial (u, v)} \sqrt{ \left( \frac{\partial (x,y)}{\partial (u',v')} \right)^2 + \left( \frac{\partial (y,z)}{\partial (u',v')} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u',v')} \right)^2 }$$

Since the factor in parentheses in front of the square root is the Jacobi determinant, we can apply the rule change of variables in multidimensional integrals to conclude that $$\int f(u,v) \sqrt{ \left( \frac{\partial (x,y)}{\partial (u,v)} \right)^2 + \left( \frac{\partial (y,z)}{\partial (u,v)} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u,v)} \right)^2 } \> du \, dv = $$ $$\int f(u',v') \sqrt{ \left( \frac{\partial (x,y)}{\partial (u',v')} \right)^2 + \left( \frac{\partial (y,z)}{\partial (u',v')} \right)^2 + \left( \frac{\partial (z,x)}{\partial (u',v')} \right)^2 } \> du' \, dv',$$ which shows that our formula gives the same answer for $\int_S f(u,v) \, d^2 A$ , no matter how we choose to parameterize $S$ .