The important observation is that th equation is homogeneous. Indeed
$\eqalign{ & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\frac{{\sqrt {{x^2} + {y^2}} }}{x} \cr & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {\frac{{{x^2} + {y^2}}}{{{x^2}}}} \cr & \frac{{dy}}{{dx}} = \frac{y}{x} - \frac{1}{h}\sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} \cr} $
Thus, let $y=vx$. Then $y'=v'x+v$; and we get
$\eqalign{ & \frac{{dv}}{{dx}}x + v = v - \frac{1}{h}\sqrt {1 + {v^2}} \cr & \frac{{dv}}{{dx}}x = - \frac{1}{h}\sqrt {1 + {v^2}} \cr & \frac{{dv}}{{\sqrt {1 + {v^2}} }} = - \frac{{dx}}{x}\frac{1}{h} \cr} $
Can you move on? The complete solution would be:
$\begin{eqnarray*} {\sinh ^{ - 1}}v = - \frac{1}{h}\left( {\log x + C} \right)\\\log \left( {v + \sqrt {1 + {v^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h}\\ \log \left( {\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} } \right) = \log \frac{1}{{\root h \of x }} - \frac{C}{h} \\\frac{y}{x} + \sqrt {1 + {{\left( {\frac{y}{x}} \right)}^2}} = \frac{k}{{\root h \of x }}\\y + \sqrt {{x^2} + {y^2}} = k{x^{1 - \frac{1}{h}}} \end{eqnarray*}$