求方程所确定的隐函数y的导数dy/dx

一楼的解法是对的,但还是可以更简化：
1、商的求导换成积的求导；
2、对积的对数求导,改成对数的和求导.
x = yln(xy) = ylnx + ylny
1 = (dy/dx)lnx + y/x + (dy/dx)lny + dy/dx
dy/dx = [1 - y/x]/[1 + ln(xy)] = y(x - y)/x(x + y)
2x²y - xy² + y³ = 0
4xy + 2x²dy/dx - y² - 2xydy/dx + 3y²dy/dx = 0
dy/dx = y(y - 4x)/(2x² - 2xy + 3y²)

关键：y是x的函数。
(y-xy')/y^2=1/xy*(y+xy'
xy-y^2=(x^2+xy)y'
y'=y(x-y)/[x(x+y)]
4xy+2x^2y'-y^2-2xyy'+3y^2y'=0
y'=y(y-4x)/(2x^2-2xy+3y^2)

O客回答正确