Pfaffian Equations Solve the Pfaffian Equation below:
$$(yz-1)dx+((z-x)x)dy+(1-xy)dz=0$$
I have tried to find the determinant, and it is $0$. So it is integrable. I let $z$ to be a constant. After I'm having problems solving it.
 A: To start, I'm assuming that we're working with $x,y,z\in\Bbb R$. As a consequence, closed and exact forms are exactly the same, i.e.
$$ \text d\omega = 0 \Leftrightarrow \omega = \text d\alpha $$
for some form $\alpha$.
Now, if
$$\omega = (yz-1)\text dx + x(z-x)\text dy + (1-xy)\text dz $$
we can see that
$$ \text d\omega = -2x \ \text dy\wedge\text dz - 2y \ \text dx\wedge\text dz - 2x \ \text dx\wedge\text dy \neq  0 $$
so $\omega$ is not closed and therefore not exact. This means we need to find $\mu(x,y,z)\neq 0$ such that $\omega/\mu$ is closed, and therefore exact. Then, there will be some $\alpha$ such that $\omega/\mu = \text d \alpha$ and so $\omega = \mu\text d\alpha$. Then the integral surfaces are $\alpha  = c$.
Take $ \mu = (xy-1)^2 $, then $ \text d(\omega/\mu) = 0 $ and so we can integrate to get
$$ \alpha(x,y,z) = \frac{x-z}{xy-1} = c $$
are the integral surfaces, or
$$ z = x + c(1-xy) $$
A: Alexander gave a perfectly fine solution, but here is the general approach you should follow if the answer doesn't "leap out" at you. There are also more elegant approaches (due to Sophus Lie, of Lie group theory) if you can spot a group that leaves the differential equation invariant.
We have the Pfaffian system $\omega = 0$ (where $\omega$ is a $1$-form, here on $\Bbb R^3$). If $d\omega = 0$, then we know from multivariable calculus how to find a potential function $f$ so that $df=\omega$. If $d\omega\ne 0$ but $\omega\wedge d\omega = 0$, then the differential equation is in fact integrable and this means that there is an integrating factor $\mu$: Namely, there is nonvanishing function $\mu$ so that $d(\mu\omega)=0$.  Since
$$d(\mu\omega) = \mu\,d\omega + d\mu\wedge\omega,$$
we want to solve
$$d\omega + \frac{d\mu}{\mu}\wedge\omega = 0,\quad\text{or}\quad
d(\log\mu)\wedge\omega = -d\omega.$$
Comparing the $dx\wedge dy$, $dz\wedge dx$, and $dy\wedge dz$ coefficients on both sides will give you a linear system of equations for the partial derivatives of $\log\mu$, and you can use standard linear algebra to solve a $3\times 3$ system of linear equations for those partial derivatives. Since we are guaranteed a solution, we will be able to integrate $\partial(\log\mu)/\partial x = P$, $\partial(\log\mu)/\partial y = Q$, $\partial(\log\mu)/\partial z = R$ for the function $\log\mu$.
EDIT: With thanks to @AntonioJPan, the error in one of the equations has been corrected.
In this particular problem, the equations are (letting $g=\log\mu$ because I'm getting tired of typing)
\begin{align*}
x(z-x)\frac{\partial g}{\partial x} +(1-yz)\frac{\partial g}{\partial y}  & =2x \\
(1-xy)\frac{\partial g}{\partial x} +(1-yz)\frac{\partial g}{\partial z} &= 2y \\
(1-xy)\frac{\partial g}{\partial y} + x(x-z)\frac{\partial g}{\partial z} &= 2x,
\end{align*}
and standard linear algebra gives us a particular solution
$$\frac{\partial g}{\partial x} = \frac {2y}{xy-1}, \quad \frac{\partial g}{\partial y} = \frac {2x}{xy-1}, \quad \frac{\partial g}{\partial z}=0.$$
Wasn't that convenient? Now we immediately get $g(x,y,z)=\log(xy-1)^2$, so $\mu(x,y,z)=(xy-1)^2$.
However, as @Antonio suggested, if we look at the general solution of this system of equations, we see that $dg$ will need to be a functional multiple of our original $1$-form $\omega$. In hindsight, this is obvious, as adding any functional multiple $\lambda\omega$ to $dg$ contributes $\lambda\omega\wedge\omega = 0$ to the equation. Indeed, this is just back to the original problem. However, since $\omega$ is nonvanishing off the curve $x=z=1/y$, we can argue that any other solution $h$ of $dh=\lambda\omega$ must satisfy $dh\wedge dg=0$, and so $h$ is a function of $g$.
