$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$
$x d y-y d x=\sqrt{x^{2}+y^{2}} d x$\
$\Rightarrow x d y=\left[y+\sqrt{x^{2}+y^{2}}\right] d x$
$\frac{d y}{d x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$ ...(1)
Let $F(x, y)=\frac{y+\sqrt{x^{2}+y^{2}}}{x}$
$\therefore F(\lambda x, \lambda y)=\frac{\lambda x+\sqrt{(\lambda x)^{2}+(\lambda y)^{2}}}{\lambda x}=\frac{y+\sqrt{x^{2}+y^{2}}}{x}=\lambda^{0} \cdot F(x, y)$
Therefore, the given differential equation is a homogeneous equation.
To solve it, we make the substitution as:
y = vx
$\Rightarrow \frac{d}{d x}(y)=\frac{d}{d x}(v x)$
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
Substituting the values of $v$ and $\frac{d y}{d x}$ in equation (1), we get:
$v+x \frac{d v}{d x}=\frac{v x+\sqrt{x^{2}+(v x)^{2}}}{x}$
$\Rightarrow v+x \frac{d v}{d x}=v+\sqrt{1+v^{2}}$
$\Rightarrow \frac{d v}{\sqrt{1+v^{2}}}=\frac{d x}{x}$
Integrating both sides, we get:
$\log \left|v+\sqrt{1+v^{2}}\right|=\log |x|+\log \mathrm{C}$
$\Rightarrow \log \left|\frac{y}{x}+\sqrt{1+\frac{y^{2}}{x^{2}}}\right|=\log |\mathrm{C} x|$
$\Rightarrow \log \left|\frac{y+\sqrt{x^{2}+y^{2}}}{x}\right|=\log |\mathrm{C} x|$
$\Rightarrow y+\sqrt{x^{2}+y^{2}}=\mathrm{C} x^{2}$
This is the required solution of the given differential equation.