start by differentiating both sides w.r.t x
#therefore d/dx (-y) = d/dx (xy + 2e^(y)sqrt(x-y))#
lets look at this term by term,
RHS,
the derivative of xy with respect to x using the product rule comes as #d/dx (x) * y + x*d/dx (y)# which is equal to #y + x dy/dx#
we need to use the chain rule and the product rule for
differentiating #2e^y sqrt(x-y)#
= #d/dx (2e^y)*sqrt(x-y) + 2e^y*d/dx (sqrt(x-y))#
the derivative of #2e^y# w.r.t x is #2e^y dy/dx#
and the derivative of #sqrt(x-y) = (x-y)^(1/2)# using the chain rule comes as #1/(2sqrt(x-y))*(1-dy/dx)#
therefore the derivative of the second term is
#2e^y dy/dx*sqrt(x-y)+(2e^y)/(2sqrt(x-y))*(1-dy/dx)#
= #2e^y sqrt(x-y)dy/dx + e^y/(sqrt(x-y)) - e^y/(sqrt(x-y)) dy/dx#
therefore, the entire equation becomes
#-dy/dx = y + x dy/dx + 2e^y sqrt(x-y)dy/dx + e^y/(sqrt(x-y)) - e^y/(sqrt(x-y)) dy/dx#
multiply both the sides by -1 to make the left hand side positive,
#=dy/dx = -y - x dy/dx - 2e^y sqrt(x-y)dy/dx - e^y/(sqrt(x-y)) + e^y/(sqrt(x-y)) dy/dx#
Now shift all the dy/dx terms to the left side and factorise
#dy/dx + x dy/dx + 2e^y sqrt(x-y)dy/dx - e^y/(sqrt(x-y)) dy/dx =
-y - e^y/(sqrt(x-y))#
#dy/dx (1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y))) = -y - e^y/(sqrt(x-y)#
therefore,
#dy/dx = (-y - e^y/sqrt(x-y)) /(1 + x + 2e^y sqrt(x-y) - e^y/(sqrt(x-y)))#