MHB Finding region of xy plane for which differential equation has a unique solution

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To find a region in the xy-plane for the differential equation x(dy/dx) = y that ensures a unique solution passing through a point (x_0, y_0), it is essential to define a rectangular region R containing (x_0, y_0). The conditions for uniqueness require that both f(x,y) and its partial derivative with respect to y are continuous within this region. The equation can be simplified to y' = f(x,y) = y/x, which is valid for x > 0 and x < 0. In both cases, the continuity of f and its partial derivative confirms the existence of unique solutions in these regions. Thus, the regions x > 0 and x < 0 are valid for unique solutions to the initial value problem.
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Determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point [math](x_0, y_0)[/math] in the region.

[math]x \frac{dy}{dx} = y[/math]

What does an xy-plane have to do with anything? I looked up the definition of unique solutions and here it is

Let R be a rectangular region in the xy-planed defined by a <=x<=b, c<=y<=d that contains the point [math](x_0, y_0)[/math] in its interior. If f(x,y) and [math]\frac{\partial{d} f}{\partial{d} y}[/math] are continuous on R then there exists some interval [math]I_0: (x_0-h, x_0+h), h>0[/math] contained in [a/b] and a unique function y(x) defined on [math]I_0[/math] that is a solution of the initial value problem.

That's a bit difficult to digest. How do I proceed?
 
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In each of the regions $D\equiv x>0$ and $D'\equiv x<0$ the differential equation is equivalent to $y'=f(x,y)=\dfrac{y}{x},$ and in both regions, $f$ and $\dfrac{\partial f}{\partial y}=\dfrac{1}{x}$ are continuous, so and according to a well known theorem, $D$ and $D'$ are solutions to your question.
 

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