f(x,y)=(x^2+y^2sin(1/(x^2+y^2)),当x^2+y^2>0时,
f(0,0)=0.
容易验证：af/ax(0,0)=0,af/ay(0,0)=0,于是f(x,y)在(0,0)可微.
但af/ax＝2xsin(1/(x^2+y^2))－2xcos(1/(x^2+y^2))/(x^2+y^2),af/ax在（0,0）不连续.