f'x=(y·(x+y^2)-xy)/(x+y^2)²=y³/(x+y^2)²,则f'x（1,1）= 1/4
fy=(x·(x+y^2)-(xy)·2y)/(x+y^2)²=(x²-xy^2)/(x+y^2)²,则f'y（1,1）= 0