Solve $y'=-\frac{x+y-2}{x-y+4}$ 
$$y'=-\frac{x+y-2}{x-y+4}$$

I have used the substation:
$x=u-1$
$y=v+3$
And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}$
And got:
$\frac{du}{u}\frac{1-z}{z^2-2z-1}$
$$u=e^{\frac{-z+2z+1}{4}+C}$$
What should I do next?
 A: $$\frac { dv }{ du } =\frac { -1-\frac { v }{ u }  }{ 1-\frac { v }{ u }  } \\ z=\frac { v }{ u } \\ v=uz\\ z+{ u }z^{ \prime  }=\frac { -1-z }{ 1-z } =\frac { 1+z }{ z-1 } \\ u{ z }^{ \prime  }=\frac { 1+z }{ z-1 } -z=\frac { 1+z-{ z }^{ 2 }+z }{ z-1 } =\frac { -{ z }^{ 2 }+2z+1 }{ z-1 } =\frac { { z }^{ 2 }-2z-1 }{ 1-z } \\ \int { \frac { 1-z }{ { z }^{ 2 }-2z-1 }  } dz=\int { \frac { du }{ u }  } \\ \int { \frac { d\left( { z }^{ 2 }-2z-1 \right)  }{ { z }^{ 2 }-2z-1 }  } =-2\int { \frac { du }{ u }  } \\ \ln { \left| { z }^{ 2 }-2z-1 \right|  } =-2\ln { \left| u \right|  } \\ { z }^{ 2 }-2z-1=\frac { C }{ { u }^{ 2 } } \\ \frac { { v }^{ 2 } }{ { u }^{ 2 } } -2\frac { v }{ u } -1=\frac { C }{ { u }^{ 2 } } \\ { v }^{ 2 }-2vu-{ u }^{ 2 }=C\\ \\  $$
then plug  substition  $x$ and $y$
A: You could approach it this way:
$$
\frac{dy}{dx} = -\frac{x + y - 2}{x - y + 4} \\
\implies x\ dy - y\ dy + 4\ dy = -x\ dx - y \ dx - 2 \ dx \\
\implies (x \ dy + y \ dx) - x \ dx - 2 \ dx - y \ dy = 0 \\
\implies \int d(xy) - \int x \ dx - 2\int dx - \int y \ dy = 0 \\
\implies x^2 + y^2 - 2xy + 4x + c = 0
$$
which is a parabola.
