Find dy/dx for 5x^2-2xy+7y^2=0

Question

I'm getting confused with signs.

Answer 1

Most likely it is for the signs in -2xy
perhaps if you look at it as +(-2x)(y)

5x^2 + (-2x)(y) + 7y^2 = 0

10x + (-2x)(dy/dx) + y(-2) + 14y dy/dx = 0
dy/dx(14y - 2x) = 2y - 10x
dy/dx = (2y - 10x)/(14y - 2x)

= (y - 5x)/(7y - x)

Answer 2

or, you can think of it as
5x^2-(2xy)+7y^2=0
and since

(2xy)' = 2y+2xy', that gives

10x-(2y+2xy')+14yy'

and go from there.