I tried to solve:
$$(\frac{x}{y} + 1 )dx + (\frac{x}{y} -1 )dy = 0$$
$(y+1)dx+(x+1)dy=0$
$M(x, y) = \frac{x}{y} + 1$
$N(x, y) = \frac{x}{y} -1$
Then, we can multiply both $M$ and $N$ by $y$ to get:
$yM(x, y) = M^{*}(x, y) = x + y$
$yN(x, y) = N^{*}(x, y) = x -y$
$\frac {\partial M^{*}}{\partial y}=1=\frac {\partial N^{*}}{\partial x}$
Hence, the differential equation is exact.
We take: $\phi_x = x+y$
We integrate with resect to $x$ and get: $\frac{x^2}{2} +xy + h(y)$
Then we differentiate with respect to $y$ and get: $\phi_y = x + h'(y)$ which is equal to $x-y$
Hence: $x + h'(y) = x-y \implies h'(y) = -y$
Integrating both sides with respect to $y$: $h(y) = -\frac{y^2}{2}$.
Our solution:
$\frac{x^2}{2} +xy -\frac{y^2}{2} = C$
Is that correct?