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?

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.
