First, you should take the derivative of on 3=(1-y)/x^2 + xy
d/dx3=d/dx( 1-y)/x^2 + d/dxxy
Use the Quotient Rule on ( 1-y)/x^2
Quotient Rule: f(x)/g(x) = (f'(x)g(x) - g'(x)f(x))/(g(x))^2
(f'(x)g(x) - g'(x)f(x))/(g(x))^2 = ((-y') * x^2 - 2x * (1-y))/(x^2)^2
It is -y' because d/dx (1-y) Derivative of any constant is 0, and the derivative of -y, using d/dx is dy/dx, which is -y'.
Use Product Rule for d/dx x*y
Product Rule: f(x)*g(x) = f'(x)*g(x) + g'(x)*f(x)
f'(x)*g(x) + g'(x)*f(x) = d/dx x*y + d/dx y *x
d/dx x*y + d/dx y *x = 1 *y + y' *x = y +y'x
Now plug in:
d/dx3=d/dx( 1-y)/x^2 + d/dxxy
0=((-y') * x^2 - 2x * (1-y))/(x^2)^2 + y +y'x
0=(-x^2y' - 2x+2xy)/(x^4) + y +y'x
0=(-y')/x^2 - 2/x^3 + (2y)/x^3 + y +y'x
0=(-y')/x^2 - (2(1 + y))/x^3 + y +y'x
Let's get the denominator to x^3
0=(-xy')/x^3 - (2 + 2y)/x^3 + (x^3y)/x^3 +(y'x^4)/x^3
0=(-xy')/x^3 + (y'x^4)/x^3- (2 + 2y)/x^3 + (x^3y)/x^3
0=(-xy' + y'x^4- 2 + 2y + x^3y)/x^3
0=(y'(-x + x^4)- 2 + y(2 + x^3))/x^3
0=y'(-x + x^4)- 2 + y(2 + x^3)
2 - y(2 + x^3)=y'(-x + x^4)
(2 - y(2 + x^3))/(-x + x^4)=y'
