\[ -\sqrt {x^2+y(x)^2}+x y'(x)-y(x)=0 \] ✓ Mathematica : cpu = 0.111233 (sec), leaf count = 13
\[\{\{y(x)\to x \sinh (\log (x)+c_1)\}\}\] ✓ Maple : cpu = 0.045 (sec), leaf count = 27
\[\frac {\sqrt {y \relax (x )^{2}+x^{2}}}{x^{2}}+\frac {y \relax (x )}{x^{2}}-c_{1} = 0\]
Hand solution
\[ xy^{\prime }=\sqrt {x^{2}+y^{2}}+y \]
Let \(y=xv\), then \(y^{\prime }=v+xv^{\prime }\) and the above becomes
\begin {align*} x\left (v+xv^{\prime }\right ) & =\sqrt {x^{2}+\left (xv\right ) ^{2}}+xv\\ x\left (v+xv^{\prime }\right ) & =x\sqrt {1+v^{2}}+xv\\ \left (v+xv^{\prime }\right ) & =\sqrt {1+v^{2}}+v\\ xv^{\prime } & =\sqrt {1+v^{2}} \end {align*}
Separable.
\[ \frac {dv}{\sqrt {1+v^{2}}}=\frac {1}{x}dx \]
Integrating
\begin {align*} \operatorname {arcsinh}\relax (v) & =\ln x+C\\ v & =\sinh \left (\ln x+C\right ) \end {align*}
Since \(y=xv\) then
\[ y=x\sinh \left (\ln x+C\right ) \]
Verification