答：
y-cos(x^2+y^2)=x
两边对x求导：y'+sin(x^2+y^2)*(2x+2yy')=1
[2ysin(x^2+y^2)+1]*y'=1-2xsin(x^2+y^2)
y'=[1-2xsin(x^2+y^2)]/[2ysin(x^2+y^2)+1]
所以：
dy/dx=[1-2xsin(x^2+y^2)]/[2ysin(x^2+y^2)+1]