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设z=（x-1）/[(x-1）^2 -y]
=(x-1)/(x²-2x+1-y)
则∂z/∂x
=[1*(x²-2x+1-y)-(x-1)(2x-2)]/(x²-2x+1-y)²
=(-x²+2x-1-y)/(x²-2x+1-y)²
∂z/∂y
=-(x-1)/(x²-2x+1-y)²*(x²-2x+1-y)'
=(x-1)/(x²-2x+1-y)²
=(x-1)/(x²-2x+1-y)
则∂z/∂x
=[1*(x²-2x+1-y)-(x-1)(2x-2)]/(x²-2x+1-y)²
=(-x²+2x-1-y)/(x²-2x+1-y)²
∂z/∂y
=-(x-1)/(x²-2x+1-y)²*(x²-2x+1-y)'
=(x-1)/(x²-2x+1-y)²
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