记：z = f(x,y) = (x^2+y^2)e^[(x^2+y^2)/xy] = u(x,y) e^[u(x,y)/v(x,y)]其中：u(x,y) = (x^2+y^2)；v(x,y) = xy；全微分：dz = df(x,y) = [∂f(x,y)/∂x] dx + [∂f(x,y)/∂y] dy ∂f(x,...∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^2+y^2) (2x-(x^2+y^2))/(x^2y)] ∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (x^2+y^2) (2y-(x^2+y^2))/(xy^2)] 不应该是 ∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^2+y^2) (2x^2-(x^2+y^2))/(x^2y)] 吗？∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (x^2+y^2) (2y^2-(x^2+y^2))/(xy^2)]我检查了一下，您说的对，应该是：∂f(x，y)/∂x = e^[(x^2+y^2)/xy] [2x + (x^4 - y^4)/(x^2y)]∂f(x，y)/∂y = e^[(x^2+y^2)/xy] [2y + (y^4 - x^4)/(xy^2)]dz = e^[(x^2+y^2)/xy] {[2x + (x^4 - y^4)/(x^2y)] dx + [2y + (y^4 - x^4)/(xy^2)] dy }