Problems with an exact differential eqution Consider the following differential equation
$$
\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx=\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy
$$ 
I want  to find its general solution. I get that this equation is exact but, I trying to solve it for this method and it seems not  work. Can someone give me a hit? 
 A: We have an exact differential equation in the form $Mdx +Ndy = 0$, with 
$$M \equiv \frac{1}{x} - \left(\frac{y}{x-y}\right)^2, N \equiv \left(\frac{x}{x-y}\right)^2 - \frac{1}{y}.$$
If we let the implicit solution be $F(x, y) = c$, since the equation is exact, it is known that $F_x = M, F_y = N.$ 
$$F_x = M \Rightarrow, F = \int \left ( \frac{1}{x} - \left(\frac{y}{x-y}\right)^2 \right )dx,$$
$$\Rightarrow F = \ln|x| + \frac{y^2}{x-y} + \phi(y),$$
where $\phi(y)$ is a function of y.
$F_y = N,$
$$\Rightarrow \frac{y(2x-y)}{(x-y)^2} +\phi'(y) = \frac{x^2}{(x-y)^2} - \frac{1}{y},$$
$$\Rightarrow \phi'(y) = \frac{x^2-2xy + y^2}{(x-y)^2} - \frac{1}{y},$$
$$\Rightarrow \phi'(y) = 1 - \frac{1}{y},$$
$$\phi(y) = y - \ln|y|+C.$$
Thus, the implicit solution is $$F(x, y) = \ln\left|\frac{x}{y}\right| + \frac{xy}{x-y}  = c.$$
A: The differential is indeed exact
$$\left(\frac{1}{x}-\frac{y^2}{(x-y)^2}\right)dx-\left(\frac{1}{y}-\frac{x^2}{(x-y)^2}\right)dy=0$$
Note that
$$\frac {dx}x=d\ln (x)$$
$$\frac {dy}y=d\ln (y)$$
And also that
$$
\begin{align}
E=&-\frac{y^2}{(x-y)^2}dx+\frac{x^2}{(x-y)^2}dy \\
E=&\frac{-y^2dx+x^2dy}{(x-y)^2}\\
E=&\frac{-y^2dx+x^2dy}{x^2y^2}\frac {(xy)^2}{(x-y)^2}\\
E=&(d(\frac 1x- \frac 1y))\frac {(xy)^2}{(x-y)^2}\\
E=&(\frac {xy}{y-x})^2d(\frac {y-x}{xy})  \\
\end{align}
$$
$$ \text {Since we have } \frac {dv}{v^2}=-d\left(\frac 1v  \right ) \implies E=-d(\frac {xy}{y-x})$$
Therefore we have
$$d  \ln x -d \ln y +d(\frac {xy}{x-y})=0$$
$$\boxed{\ln (\frac xy)+\frac {xy}{x-y}=K}$$
