\[ -x (y(x)-x) \sqrt {x^2+y(x)^2}+x y'(x)-y(x)=0 \] ✓ Mathematica : cpu = 0.30524 (sec), leaf count = 221
\[\left \{\left \{y(x)\to \frac {x-2 \sqrt {x^2 \tanh ^2\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )-x^2 \tanh ^4\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )}}{-1+2 \tanh ^2\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )}\right \},\left \{y(x)\to \frac {x+2 \sqrt {x^2 \tanh ^2\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )-x^2 \tanh ^4\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )}}{-1+2 \tanh ^2\left (\frac {1}{2} \left (-\sqrt {2} x^2-2 \sqrt {2} c_1\right )\right )}\right \}\right \}\] ✓ Maple : cpu = 0.221 (sec), leaf count = 49
\[\left \{\frac {\sqrt {2}\, x^{2}}{2}-c_{1}-\ln \left (x \right )+\ln \left (\frac {2 \left (x +y \left (x \right )+\sqrt {2 x^{2}+2 y \left (x \right )^{2}}\right ) x}{-x +y \left (x \right )}\right ) = 0\right \}\]
\[ xy^{\prime }=x\left ( y-x\right ) \sqrt {y^{2}-x^{2}}+y \]
Let \(y=xu\), then \(y^{\prime }=u+xu^{\prime }\) and the above becomes
\begin {align*} x\left ( u+xu^{\prime }\right ) & =x\left ( xu-x\right ) \sqrt {\left ( xu\right ) ^{2}-x^{2}}+xu\\ \left ( u+xu^{\prime }\right ) & =\left ( xu-x\right ) \sqrt {\left ( xu\right ) ^{2}-x^{2}}+u\\ xu^{\prime } & =\left ( xu-x\right ) x\sqrt {u^{2}-1}\\ u^{\prime } & =x\left ( u-1\right ) \sqrt {u^{2}-1} \end {align*}
Separable.
\begin {align*} \frac {du}{\left ( u-1\right ) \sqrt {u^{2}-1}} & =xdx\\ \frac {-u-1}{\sqrt {u^{2}-1}} & =\frac {x^{2}}{2}+C \end {align*}
But \(y=xu\), hence
\[ \frac {-\frac {y}{x}-1}{\sqrt {\left ( \frac {y}{x}\right ) ^{2}-1}}=\frac {x^{2}}{2}+C \]
Let \(\frac {y}{x}=z\)
\begin {align*} \frac {-z-1}{\sqrt {z^{2}-1}} & =\frac {x^{2}}{2}+C\\ -z-1 & =\sqrt {z^{2}-1}\left ( \frac {x^{2}}{2}+C\right ) \\ \left ( -z-1\right ) ^{2} & =\left ( z^{2}-1\right ) \left ( \frac {x^{2}}{2}+C\right ) ^{2}\\ z^{2}+1+2z & =z^{2}\left ( \frac {x^{2}}{2}+C\right ) ^{2}-\left ( \frac {x^{2}}{2}+C\right ) ^{2}\\ z^{2}\left ( 1-\left ( \frac {x^{2}}{2}+C\right ) ^{2}\right ) +2z+1+\left ( \frac {x^{2}}{2}+C\right ) ^{2} & =0 \end {align*}
\(\allowbreak \)
Solving for \(z\) (quadratic formula, some conditions apply), one of the solutions is
\[ z=\frac {4Cx^{2}+4C^{2}+x^{4}+4}{4Cx^{2}+4C^{2}+x^{4}-4}\]
Hence
\[ y=x\frac {4Cx^{2}+4C^{2}+x^{4}+4}{4Cx^{2}+4C^{2}+x^{4}-4}\]
Need to work on verification. Kamke gives the final solution as
\[ y=x\frac {-2Cx^{2}+C^{2}+x^{4}+4}{-2Cx^{2}+C^{2}+x^{4}-4}\]
I am not sure where my error now is. Need to look at this again.