Term 1:
Use the chain rule:
#(d(y²))/dx = 2ydy/dx#
Term 2:
Use the chain rule:
#(d{f(u(x))})/dx = ((df(u))/(du))(du/dx)#
let #u(x) = -xy#, then #f(u) = 2e^u#
#(df(u))/(du) = (d(2e^u))/(du) = 2e^u#
Use the product rule on u to find #(du)/dx#:
#(du)/dx = -y - xdy/dx#
Substituting into the chain rule:
#(2e^u)(-y - xdy/dx)#
Reverse the substitution for u:
#(2e^(-xy))(-y - xdy/dx) = -2ye^(-xy) - 2xe^(-xy)dy/dx#
Term 3:
#(d6)/dx = 0#
Return the terms back into the equation:
#2ydy/dx -2ye^(-xy) - 2xe^(-xy)dy/dx = 0#
Move the terms not containing #dy/dx# to the right side:
#2ydy/dx - 2xe^(-xy)dy/dx = 2ye^(-xy)#
Factor out 2 and #dy/dx# on the left:
#2(y - xe^(-xy))dy/dx = 2ye^(-xy)#
Divide by the coefficient of #dy/dx#:
#dy/dx = (ye^(-xy))/(y - xe^(-xy))#
