sin(x²+y²)-xy²=0
[sin(x²+y²)]'-(xy²)'=0'
cos(x²+y²)*(x²+y²)'-[x'y²+x(y²)']=0
cosx(x²+y²)*(2xy²+2yy')-[y²+2xyy']=0
-y²[1-2xcosx(x²+y²)]+2y'y[cosx(x²+y²)+x]=0
y'={y[1-2xcosx(x²+y²)]}/{2[cosx(x²+y²)+x]}