Answer
$C=x^2+\tan (xy)$
Work Step by Step
We are given:
$$x\sec^2(xy)dy=-[y\sec^2(xy)+2x]dx$$
$$x\sec^2(xy)dy+[y\sec^2(xy)+2x]dx=0$$
We have that:
$$M(x,y)=y\sec^2(xy)+2x \wedge N(x,y)=x\sec^2(xy)$$
then
$$M'(x,y)=\sec^2(xy)+2xy\sec^2(x)\tan(xy)$$
$$N'(x,y)=\sec^2(xy)+2xy\sec^2(x)\tan(xy)$$
$$M'(x,y)=N'(x,y)$$
So this equation exists.
We have:
There exists a potential function $\phi$ satisfies:
$$\frac{\partial \phi}{\partial x}=M \wedge \frac{\partial \phi}{\partial y}=N$$
$$\frac{\partial \phi}{\partial x}=y\sec^2(xy)+2x \wedge \frac{\partial \phi}{\partial y}=x\sec^2(xy)$$
Integrating $\frac{\partial \phi}{\partial x}=y\sec^2(xy)+2x$ we get
$$\phi=x^2+\tan (xy)+f(y)$$
Since $\frac{\partial \phi}{\partial y}=x\sec^2(xy)+f'(y)$
We obtain:
$$f'(y)=0\rightarrow f(y)=C$$
$C$ is constant of integration
The potential solution is:
$$\phi=x^2+\tan (xy)$$
$$C=x^2+\tan (xy)$$
Hence general solution is $C=x^2+\tan (xy)$
