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'#
