z(x,y)=f(x^2+y^2,x-y)
dz(x,y)=f1'(2xdx+2ydy)+f2'(dx-dy)
=(2xf1'+f2')dx+(2yf1'-f2')dy
所以：
∂z/∂x=(2xf1'+f2')
∂z/∂y=(2yf1'-f2')
对方程
z = f(y/x,x+2y)
的两端求微分,得
dz = f1*[(xdy-ydx)/x²]+f2*(dx+2dy)
= [-(y/x²)f1+f2]dx+[(1/x)f1+2*f2]dy,
得到
Dz/Dx = -(y/x²)f1+f2,Dz/Dy = (1/x)f1+2*f2,
于是
D²z/DxDy = (D/Dx)(Dz/Dy)
= (D/Dx)[(1/x)f1+2*f2]
= [(-1/x²)*f1+(1/x)*[-(y/x²)f11+f12]+2*[(1/x)f21+2*f22]
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