$\Omega=\{(x,y)\in R^2|x>0\}\subset R^2$ is open set, p is a $C^1$ function that $p\colon \Omega\rightarrow R$ , $p_0=p(x_0,y_0)$. we difine function $F \colon \Omega \rightarrow R$,
$(x,y)\mapsto y-y_0+\frac{p(x,y)+p_0}{2}\big(\frac{1}{x}-\frac{1}{x_0}\big)$.
Proof there is an open neighborhood U at $p_0$ as well as the $C^1$ function $\phi$ on U, making the
$F(x,y)=0$ be equivalent to $y=\phi(x)$
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