Solve the differential equation $\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$ This is problem 9, exercise 10 from Tannebaum and Pollard's ODE book. I have deduced that the differential equation is exact, but I can't find all the integrable combinations. Any hints that would help me to move forward would be great!

Solve the differential equation  $\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$

Solution.
We have,
$\begin{align}
P&=\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\\
\frac{\partial P}{\partial y}&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{(1+x^{2}y^{2})(x-4xy)-(xy-2xy^{2})(2x^{2}y)}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{x-4xy+x^{3}y^{2}-4x^{3}y^{3}-2x^{3}y^{2}+4x^{3}y^{3}}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x}{1+x^{2}y^{2}}+\left[\frac{x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\right]\\
&=\frac{x(1+x^{2}y^{2})+x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{x+x^{3}y^{2}+x-4xy-x^{3}y^{2}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy}{(1+x^{2}y^{2})^{2}}\\
Q&=\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\\
\frac{\partial Q}{\partial x}&=\frac{(1+x^{2}y^{2})(2x-4xy)-(x^{2}-2x^{2}y)(2xy^{2})}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy+2x^{3}y^{2}-4x^{3}y^{3}-2x^{3}y^{2}+4x^{3}y^{3}}{(1+x^{2}y^{2})^{2}}\\
&=\frac{2x-4xy}{(1+x^{2}y^{2})^{2}}
\end{align}$
Since $\partial{P}/\partial{y}=\partial{Q}/\partial{x}$, this is an exact differential equation.
We know that-
$\begin{align}
d(x\cdot \arctan(xy))&=\arctan(xy)dx+x\frac{1}{1+x^{2}y^{2}}(xdy+ydx)\\
&=\arctan(xy)dx+\frac{x^{2}dy}{1+x^{2}y^{2}}+\frac{xydx}{1+x^{2}y^{2}}
 \end{align}$
I am not able to find an integrable combination for the remaining two terms.

Your solution is good, here is another approach:
$$\left(\arctan(xy)+\frac{xy-2xy^{2}}{1+x^{2}y^{2}}\right)dx+\left(\frac{x^{2}-2x^{2}y}{1+x^{2}y^{2}}\right)dy=0$$
$$\arctan(xy)dx +\frac{xy}{1+x^{2}y^{2}}dx +\frac{x^{2}}{1+x^{2}y^{2}}dy -\frac{2xy^{2}}{1+x^{2}y^{2}}dx  -\frac{2x^{2}y}{1+x^{2}y^{2}}dy=0$$
$$\arctan(xy)dx +\frac{xy\,dx+x^2\,dy}{1+x^{2}y^{2}}  -\frac{2xy^{2}\,dx+2x^{2}y\,dy}{1+x^{2}y^{2}}=0$$
$$\arctan(xy)dx +x\frac{d(xy)}{1+x^{2}y^{2}}  -\frac{2xy\,d(xy)}{1+(xy)^{2}}=0$$
$$d(x.\arctan(xy))-\frac{2xy\,d(xy)}{1+(xy)^{2}}=0$$
$$x\arctan(xy)-\ln(1+(xy)^{2})+C=0$$
