Exact Differential Equation Integrating Factor 
Finding an integrating factor can be a genuine mathematical art. 
  However, certain differential forms can remind us of differentiation
  techniques that may aid in the solution of the equation at hand.  For
  example, seeing $x\ dy + y\ dx$ reminds us of the product rule, as in
  $d(xy) = x\ dx + y\ dy$ [sic], and $x\ dy - y\ dx$ might remind us of the
  quotient rule, $d(x/y) = \frac{y\ dx - x\ dy}{y^2}$.  In the equation
  $x\ dy + y\ dx + 3xy^2\ dy = 0$, we are again reminded of the product
  rule.  In fact, if you multiply the equation by $\frac{1}{xy}$, then
  $\frac{x\ dy\ +\ y\ dx}{xy} + 3y\ dy = 0 \implies d(ln(xy)) + 3y\ dy = 0 \implies ln(xy) + \frac{3}{2}y^2 = C$.  Use these ideas to find a
  general solution for the differential equation $$(x^2 - y^2)(x\ dy + y\ dx) = 2xy(x\ dy - y\ dx)$$

I'm not really sure where to start with this.  I can rewrite the equation as $(x^2 - y^2)\ d(xy) = 2x^3y\ d(y/x)$ and solve for one of the differentials, but not in a way where I can integrate the RHS.  I noticed that $x\ dy + y\ dx$ is cleanly expressible as a dot product, and since $d\vec r$ = $\begin{bmatrix}dx \\ dy\end{bmatrix}$, integrating $d(xy)$ could be viewed as calculating work, where $\vec f(x,\ y) = \begin{bmatrix}y \\ x\end{bmatrix}$.  However, there doesn't appear to be a curve, so I can't set integration bounds or use Green's Theorem.  Multiplying everything out didn't spark any ideas either.
 A: Starting from your original differential equation of
$$(x^2 - y^2)(x\ dy + y\ dx) = 2xy(x\ dy - y\ dx) \tag{1}\label{eq1}$$
divide both sides by $x^2$ to get
$$\left(1 - \left(\frac{y}{x}\right)^2\right)(x\ dy + y\ dx) = 2xy\left(\frac{x\ dy - y\ dx}{x^2}\right) \tag{2}\label{eq2}$$
Using the suggestions for the differentials, we get
$$\left(1 - \left(\frac{y}{x}\right)^2\right)d\left(xy\right) = 2xy\left(d\left(\frac{y}{x}\right)\right) \tag{3}\label{eq3}$$
We now have a separable equation in $xy$ and $\frac{y}{x}$. Making the appropriate adjustments, i.e., dividing both sides by $xy\left(1 - \left(\frac{y}{x}\right)^2\right)$, gives
$$\frac{d\left(xy\right)}{xy} = \frac{2d\left(\frac{y}{x}\right)}{1 - \left(\frac{y}{x}\right)^2} \tag{4}\label{eq4}$$
You can now integrate both sides appropriately and simplify the results. I assume you can finish the remaining calculations yourself.
A: Dividing both sides by $xy(x^2-y^2)$, we get
$$\frac{d(xy)}{xy}=\frac{d(x+y)}{x+y}-\frac{d(x-y)}{x-y}$$
$$\ln(xy)=\ln(x+y)-\ln(x-y)+C_1$$
Answer:
$$\frac{xy(x-y)}{x+y}=C$$
