两边对x求导:
[e^(x+y)](x + y)' + 1 + 2yy' = 0
[e^(x+y)](1 + y') + 1 + 2yy' = 0
[2y + e^(x+y)]y' = -[e^(x + y) + 1]
y' = -[e^(x + y) + 1]/[2y + e^(x+y)]
x = 0, y = 0; 这个解很容易看出;但另一个不容易求,约为y = -0.71
在(0, 0)处, y' = -(e^0 + 1)/(2*0 + e^0) = -2
在(0, -0.71)处,y' = -[e^(-0.71) + 1]/[2*0 + e^(-0.71)] =1.58