If $P(x,y)$ and $Q(x,y)$ are $\mathcal{C}^1$ functions such that $\oint_C Pdx+Qdy=0$ for every circle $C$ in $\mathbb{R}^2$, then $P_y(x,y)=Q_x(x,y)$

Is the following correct?

Let $P(x,y),\, Q(x,y)$ be $\mathcal{C}^1$ functions in $\mathbb{R}^2$ such that $\oint_C Pdx+Qdy=0$ for every circle $C$ in $\mathbb{R}^2$
Suppose that there is a point $(x_0,y_0)$ such that $\frac{\partial Q}{\partial x}(x_0,y_0)\neq\frac{\partial P}{\partial y}(x_0,y_0)$
Then, $\frac{\partial Q}{\partial x}(x_0,y_0)-\frac{\partial P}{\partial y}(x_0,y_0)\neq0$
Since $P$ and $Q$ are $\mathcal{C}^1$, we have that $f(x,y):=\frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y)$ is continuous, therefore, there exist $\varepsilon>0$ such that $f(x,y)\neq0\,\forall(x,y)\in B:=B((x_0,y_0),\varepsilon)$
More than that, sgn$(f(x,y))=$sgn$(f(x_0,y_0))\,\forall (x,y)\in B$, therefore, $\int_B f(x,y) dxdy$ has the same sign as $f(x_0,y_0)$, and does not equal zero
By Green's theorem, we have that $\int_B (\frac{\partial Q}{\partial x}(x,y)-\frac{\partial P}{\partial y}(x,y))dxdy=\oint_{\partial B} Pdx+Qdy$, wich means $\oint_{\partial B} Pdx+Qdy\neq0$
This is a contradiction, since $\partial B$ is a circle
Therefore, $\frac{\partial P}{\partial y}(x,y)=\frac{\partial Q}{\partial x}(x,y)$ for every $(x,y)\in\mathbb{R}^2$

The part I'm not sure about is the part about the signs of the functions and the integral