It looks like you have the differential equation,
[tex]\dfrac{6x-7}{x^2-2x} \, dx + \dfrac{y}{y-1} \, dy = 0[/tex]
which is separable as-is, and we can rearrange terms as
[tex]\dfrac{y}{y-1} \, dy = -\dfrac{6x-7}{x^2-2x} \, dx[/tex]
Take the integral of both sides. On the left, we have
[tex]\dfrac{y}{y-1} = \dfrac{(y-1)+1}{y-1} = 1 + \dfrac1{y-1}[/tex]
so that
[tex]\displaystyle \int\frac{y}{y-1} \, dy = y + \ln|y - 1|[/tex]
On the right, we decompose into partial fractions:
[tex]-\dfrac{6x-7}{x^2-2x} = -\dfrac{6x-7}(x(x-2)} = \dfrac ax + \dfrac b{x-2}[/tex]
[tex]-\dfrac{6x-7}{x^2-2x} = \dfrac{a(x-2)+bx}{x^2-2x}[/tex]
[tex]-\dfrac{6x-7}{x^2-2x} = \dfrac{(a+b)x-2a}{x^2-2x}[/tex]
[tex]\implies\begin{cases}a+b=-6 \\ -2a=7\end{cases} \implies a=-\dfrac72 \text{ and } b = -\dfrac52[/tex]
[tex]\implies -\dfrac{6x-7}{x^2-2x} = -\dfrac12 \left(\dfrac7x + \dfrac5{x-2}\right)[/tex]
so that
[tex]\displaystyle \int -\frac{6x-7}{x^2-2x} \, dx = -\frac72 \ln|x| - \frac52 \ln|x-2|[/tex]
So, upon integrating both sides we get
[tex]y + \ln|y - 1| = -\dfrac72 \ln|x| - \dfrac52 \ln|x - 2| + C[/tex]
[tex]y + \ln|y - 1| = -\dfrac12 \left(7\ln|x| + 5 \ln|x - 2|\right) + C[/tex]
[tex]y + \ln|y - 1| = -\dfrac12 \left(\ln|x|^7 + \ln|x - 2|^5\right) + C[/tex]
[tex]\boxed{y + \ln|y - 1| = -\dfrac12 \ln\left|x^7(x-2)^5\right| + C}[/tex]
