\[ -x \sqrt {x^2+y(x)^2}+x y'(x)-y(x)=0 \] ✓ Mathematica : cpu = 0.01981 (sec), leaf count = 12
\[\left \{\left \{y(x)\to x \sinh \left (c_1+x\right )\right \}\right \}\]
✓ Maple : cpu = 2.378 (sec), leaf count = 28
\[ \left \{ \ln \left ( \sqrt { \left ( y \left ( x \right ) \right ) ^{2}+{x}^{2}}+y \left ( x \right ) \right ) -x-\ln \left ( x \right ) -{\it \_C1}=0 \right \} \]
\[ xy^{\prime }=x\sqrt {x^{2}+y^{2}}+y \]
Let \(y=xv\), then \(y^{\prime }=v+xv^{\prime }\) and the above becomes
\begin {align*} x\left ( v+xv^{\prime }\right ) & =x\sqrt {x^{2}+\left ( xv\right ) ^{2}}+xv\\ \left ( v+xv^{\prime }\right ) & =x\sqrt {1+v^{2}}+v\\ xv^{\prime } & =x\sqrt {1+v^{2}}\\ v^{\prime } & =\sqrt {1+v^{2}} \end {align*}
Separable.
\[ \frac {dv}{\sqrt {1+v^{2}}}=dx \]
Integrating
\begin {align*} \operatorname {arcsinh}\left ( v\right ) & =x+C\\ v & =\sinh \left ( x+C\right ) \end {align*}
Since \(y=xv\) then
\[ y=x\sinh \left ( x+C\right ) \]
Verification
ode:=x*diff(y(x),x)=x*sqrt(x^2+y(x)^2)+y(x); y0:=x*sinh(x+_C1); odetest(y(x)=y0,ode) assuming x>0; 0