偏导为：-2xy/(x²+y²)²
解:
原式=∂z/∂x=1/(1+y²/x²)*(-y/x²)=-y/(x²+y²)
∂z/∂y=1/(1+y²/x²)*1/x=x/(x²+y²)
∂²z/∂x²=y/(x²+y²)*2x=2xy/(x²+y²)²
∂²z/∂x∂y=-[x²+y²-2y²]/(x²+y²)²=(y²-x²)/(x²+y²)²
∂²z/∂y²=-2xy/(x²+y²)²
[arctan y/x]'= 1/[1+ (y/x)^2] * (y/x)'
=x^2/[x^2+y^2]* y'/x - x^2/[x^2+y^2]* y/x^2
=x^2*y'/[x^3+xy^2] - y/[x^2+y^2]
