要有步骤啊啊，谢谢
两边对x求导:
[e^(2x - y)](2 - y') - [cos(xy)]*(y + xy') = 0
x = 0, y, = 裂空1: y' = 2[e^(-1)]/[e^(-1) + 0] = 2 - e
法线斜率指塌 = 1/(e - 2)
法线方程: y - 1 = 唯源圆(x - 0)/(e - 2)
y = x/(e - 2) + 1
另见图
解：悔搜方程两边同时对x求导得
e^(2x-y) . (2-dy/拍颤dx)-cos(xy).(y+xdy/dx)=0
将（0，1）带入求得dy/袭前败dx=2-e
所以法线的斜率k=-1/(2-e)
由此法线方程求得为：y=1-x/(2-e)
e^(2x-y)―sin(xy)=e-1求导：码顷皮
（2-y‘）e^(2x-y)-(y+xy')cos(xy)=0
带乎滑入（0,1）迟差求得y'=2-e
法线与切线垂直，它们的积为-1.发现斜率为1/(e-2)
方程：y-1=x/(e-2)
x=0=>y=ln(e-1)
e^(2x-y)―sin(xy)=e-1
=>y'=[ycosxy-2e^(2x-y)]/[e^(2x-y)-xcosxy]
=>y=f(x)在点(0,1)处的切线的斜率为：亏卜信[ln(e-1)-2(e-1)]/(e-1)=-2+ln(e-1)
=>销轮y=f(x)在点(0,1)处的法线方弊团程为:y-1=[2-ln(e-1)]x
即[2-ln(e-1)]x-+1＝0
e^(2x-y)―sin(xy)=e-1
(2-y')e^(2x-y)-(y+xy')cos(xy)=0
2e^(2x-y)-ycos(xy)=(e^(2x-y)+xcos(xy))y'
y'悉盯含=(2e^(2x-y)-ycos(xy))/(e^(2x-y)+xcos(xy))
f'(0)=(2/e-1)/(1/e)
=2-e
法线睁笑斜率=-1/则行(2-e)=1/(e-2)
y=f(x)在点(0,1)处的法线方程为 y=x/(e-2)+1