解：∂z/∂y={y'·√(x²+y²)-y[∂√(x²+y²)]/∂y}/[√(x²+y²)]²=[√(x²+y²)-y·(1/2)·2y/√(x²+y²)]/(x²+y²)=[√(x²+y²)-y²/√(x²+y²)]/(x²+y²)=[(x²+y²)-y²]/(x²+y²)³=x²/(x²+y²)³
