implicit differentation means you assume y is a function of x:
Dx(xe^y) -Dx(10x) +Dx(3y) = Dx(0)
Dx(xe^y) use the product rule:
(dx/dx)e^y + (dy/dx)xe^y = e^y + y'xe^y
Dx(10x) = (dx/dx)(10) = 10
Dx(3y) = 3(dy/dx) = 3y'
Dx(0) = (d0/dx)0 = 0
Get all your y' on one side and everything else to the other:
e^y + y'xe^y -10 +3y' = 0
y'xe^y + 3y' = 10 -e^y
y'(xe^y +3) = 10 -e^y
(dy/dx) =y' =(10-e^y) / (xe^y +3)
