$\def\d{\mathrm{d}}$I'm thinking about this differential equation
$$\frac{3}{2} x\,\d x + \frac{x}{y}\,\d y = 0.$$
If functions $P(x,y), Q(x,y)$ are difined as$$P = \frac{3}{2} x,\ Q = \frac{x}{y},$$ this differential equation can be rewritten as
$$P\,\d x + Q\,\d y = 0.$$
By differentiating $P$ and $Q$, one can easily get
$$\frac{\partial P}{\partial y} = 0,\ \frac{\partial Q}{\partial x} = \frac{1}{y} \Longrightarrow \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}.$$
According to Green's theorem,
$$\oint _C (P\,\d x + Q\,\d y) = \iint _R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \,\d x \d y.$$
Now $P\,\d x + Q\,\d y = 0$, so that $\oint _C (P\,\d x + Q\,\d y)$ is always zero, and thus$$\iint _R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \,\d x \d y$$ is always zero. This results in $$\frac{\partial Q}{\partial x} = \frac{\partial P}{\partial y}.$$
Why is my way of using Green's theorem incorrect?
I have always maintained that the notation $\oint_C Pdx+Qdy$ for the line integral of a vector field is a really bad one, and this is a good example.
The differential equation $Pdx+Qdy=0$ is satisfied when $y$, as a function of $x$, is the right solution (namely, $y(x)=ce^{-3x/2}$).
In the line integral $\oint_C Pdx+Qdy$ there is nothing of that, so it makes no sense to say that $Pdx+Qdy=0$. You can see it by trying some parametrization. Note that we need to avoid zero, because if we allow $y$ to be zero, Green does not apply (continuity of the derivative will not be satisfied). Let's try with the circle of radius $1$ centered at $(0,2)$. So we parametrize $x(t)=\cos t$, $y(t)=2+\sin t $, $0\leq t\leq2\pi$. Then \begin{align} \oint_C Pdx+Qdy&=\int_0^{2\pi}\left(\frac32\,x(t)x'(t)+\frac{x(t)}{y(y)}y'(t)\right)\,dt\\ \ \\ &=\int_0^{2\pi}\left(-\frac{3\cos t\,\sin t}{2}+\frac{\cos^2t}{2+\sin t}.\right)\,dt \end{align} You can clearly see that, contrary to what you stated, the expression between brackets is not zero.