[tex](x^2 - y^2) \, dx + 2xy \, dy = 0[/tex]
Multiply both sides by [tex]\frac1{x^2}[/tex].
[tex]\left(1 - \dfrac{y^2}{x^2}\right) \, dx + \dfrac{2y}x \, dy = 0[/tex]
Substitute [tex]y=vx[/tex], so [tex]v=\frac yx[/tex] and [tex]dy=x\,dv+v\,dx[/tex].
[tex](1-v^2) \, dx + 2v (x\,dv + v\,dx) = 0[/tex]
[tex](1 + v^2) \, dx + 2xv \, dv = 0[/tex]
Separate the variables.
[tex]2xv\,dv = -(1 + v^2) \, dx[/tex]
[tex]\dfrac{v}{1+v^2}\,dv = -\dfrac{dx}{2x}[/tex]
Integrate both sides
[tex]\displaystyle \int \frac{v}{1+v^2}\,dv = -\frac12 \int \frac{dx}x[/tex]
On the left side, substitute [tex]w=1+v^2[/tex] and [tex]dw=2v\,dv[/tex].
[tex]\displaystyle \frac12 \int \frac{dw}w = -\frac12 \int\frac{dx}x[/tex]
[tex]\displaystyle \ln|w| = -\ln|x| + C[/tex]
Solve for [tex]w[/tex], then [tex]v[/tex], then [tex]y[/tex].
[tex]e^{\ln|w|} = e^{-\ln|x| + C}[/tex]
[tex]w = e^C e^{\ln|x^{-1}|}[/tex]
[tex]w = Cx^{-1}[/tex]
[tex]1 + v^2 = Cx^{-1}[/tex]
[tex]1 + \dfrac{y^2}{x^2} = Cx^{-1}[/tex]
[tex]\implies \boxed{x^2 + y^2 = Cx}[/tex]
Your mistake is in the first image, between third and second lines from the bottom. (It may not be the only one, it's the first one that matters.)
You incorrectly combine the fractions on the left side.
[tex]\dfrac1{-2v} -\dfrac v{-2} = \dfrac1{-2v} - \dfrac{v^2}{-2v} = \dfrac{1-v^2}{-2v} = \dfrac{v^2-1}{2v}[/tex]
