∂z/∂x=1/√(1-x²y²)*(xy)'=y/√(1-x²y²)=y*(1-x²y²)^(-1/2)所以∂²z/∂x²=-1/2*y*(1-x²y²)^(-3/2)*(1-x²y²)=-1/2*y/[(1-x²...