设由下列方程确定y是x的函数，求dy/dx（其中a、b为常数）

(1)x^2+y^2-xy=1

(2)y^2-2axy+b=0

(3)xy-sin(πy^2)=0,求y' | x=0.y=1

请教详解，谢谢~

(1)2x+2y*dy/dx-y-xdy/dx=0

即dy/dx=(y-2x)/(2y-x)

(2)2y*dy/dx-2a(y+xdy/dx)=0

即dy/dx=2ay/(2y-2ax)

(3)y+xdy/dx-cos(πy²)*2y*dy/dx=0

即dy/dx=-y/[x-2ycos(πy²)]

(1),2x+2yy'-y-xy'=0,

y'=dy/dx=(y-2x)/(2y-x);

(2),2yy'-2ay-2axy'=0,

y'=dy/dx=ay/(y-ax);

(3),y+xy'-[cos(πy²)]*2πyy'=0,

y'=y/[2πycos(πy²)-x]=-1/(2π),(x=0,y=1)