(1) z=xsin(y/x), z'<x>=sin(y/x)-(y/x)cos(y/x), z'<y>=cos(y/x),
(2) z=√ln(xy), z'<x>=1/[2x√ln(xy)], z'<y>=1/[2y√ln(xy)].
(3) z=(1+xy)^y, z'<x>=y^2(1+xy)^(y-1),
lnz=yln(1+xy), z'<y>/z=ln(1+xy)+xy/(1+xy),
z'(y>=(1+xy)^y*[ln(1+xy)+xy/(1+xy)].
(1) z=arctan(y/x), z'<x>=-y/(x^2+y^2), z'<y>=x/(x^2+y^2).
(2) z=√(3x^2+y^2), z'<x>=3x/√(3x^2+y^2), z'<y>=y/√(3x^2+y^2).
f(x,y)=x^2+(y-1)arcsin√(x/y),
f'<x>(x,y)=2x+(y-1)/√(y^2-x^2),
f'<x>(x,1)=2x.或
f(x,1)=x^2, 得 f'<x>(x,1)=2x。u'<x>=f'<r>r'<x>=[x/√(x^2+y^2)]f'(r)
u''<xx>=[y^2/(x^2+y^2)^(3/2)]f'(r)+[x/√(x^2+y^2)]^2*f''(r)
=[y^2/(x^2+y^2)^(3/2)]f'(r)+[x^2/(x^2+y^2)]*f''(r);
由轮换性，得
u''<yy>=[x^2/(x^2+y^2)^(3/2)]f'(r)+[y^2/(x^2+y^2)]*f''(r)。
则 u''<xx>+u''<yy>=f''(r)+f'(r)/√(x^2+y^2)
=f''(r)+f'(r)/r，由题设得
f''(r)+f'(r)/r=4 为f'(r)的一阶线性微分方程， 则
f'(r)=e^(-∫dr/r)[∫4e^(∫dr/r)dr+C1] = (1/r)[∫4rdr+C1]
= (1/r)(2r^2+C1) = 2r+C1/r,
得 f(r)=r^2+C1lnr+C2
