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)))#