y³+3y²xy'=y'+1y'=(y³-1)/(1-3xy²)①y"=[(y³-1)/(1-3xy²)]'=[3y²y'(1-3xy²)+(3y²+3y²xy')(y³-1)]/(1-3xy²)²②把①代入②得到y"=6y(y³-1)(y-x-2xy³)/(1-3xy²)³ 绝对原创！请加分！
