2x(e^(-2y) ) + x^2(e^(-2y) )(-2dy/dx)= 1/(xy) (xdy/dx + y)
2x e^(-2y) -2y x^2 e^(-2y) = (1/y) dy/dx + 1/x
2x e^(-2y) - 1/x = dy/dx (1/y + 2x^2 e(-2y) )
dy/dx = ( 2x e^(-2y) - 1/x ) / ((1/y + 2x^2 e(-2y)) )
still pretty messy looking, I don't know how far you have to take it.
Also, you better check my work. It is always harder to just type it here, rather than do it on paper.
Suppose (x^2)*(e^(-2y))= ln(xy)
Find dy/dx by implicit differentiation.
1 answer