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))#