$y'=\frac{x-y+2}{x+y-1}$ First, I substitute $x=\xi + h$, $y=\eta+k$. The equation becomes $$y'=\frac{\xi+h-(\eta+k)+2}{\xi+h+\eta+k-1}$$
and this becomes homogenous if we let $h=-\frac{1}{2}$ and $k=\frac{3}{2}$ ( We solve the constant terms such that they become zero ). I think it's correct this far, but what next? At this point we'd have $$\eta'=\frac{\xi-\eta}{\xi+\eta}$$
which is homogenous, and I tried substituting $\eta = u\xi$. This would lead to the equation $$u'\xi+u=\frac{1-u}{1+u}$$
I reduce $u$ from both sides. This will give $$u'\xi=\frac{1-u}{1+u}-u=\frac{1-2u-u^2}{1+u}$$
or $$\frac{du}{d\xi}\xi=\frac{1-2u-u^2}{1+u}$$
inverting this gives the relation $$\frac{1}{\xi}d\xi= \frac{1+u}{1-2u-u^2}du$$
So there has been some mistake made. The correct answer does not involve logarithms, and $\ln(\xi)$ will inevitably be a part of the solution. Can somebody spot my mistake?

Just take expinential on both sides. Here is another way:
$$y'=\frac{x-y+2}{x+y-1}$$
$$y'\color {red} {+1}=\frac{x-y+2}{x+y-1} \color {red} {+1}$$
$$(y+x-1)'=\frac{2x+1}{x+y-1}$$
$$(y+x-1)d(y+x-1)=( {2x+1})dx$$
Integrate both sides:
$$(y+x-1)^2=2x(x+1)+C$$
