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The given differential equation is:

$\left(x^{2}-y^{2}\right) d x+2 x y d y=0$

$\Rightarrow \frac{d y}{d x}=\frac{-\left(x^{2}-y^{2}\right)}{2 x y}$                ...(1)

Let $F(x, y)=\frac{-\left(x^{2}-y^{2}\right)}{2 x y}$.

$\therefore F(\lambda x, \lambda y)=\left[\frac{(\lambda x)^{2}-(\lambda y)^{2}}{2(\lambda x)(\lambda y)}\right]=\frac{-\left(x^{2}-y^{2}\right)}{2 x y}=\lambda^{0} \cdot F(x, y)$

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 $y$ and $\frac{d y}{d x}$ in equation (1), we get:

$v+x \frac{d v}{d x}=-\left[\frac{x^{2}-(v x)^{2}}{2 x \cdot(v x)}\right]$

$v+x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}$

$\Rightarrow x \frac{d v}{d x}=\frac{v^{2}-1}{2 v}-v=\frac{v^{2}-1-2 v^{2}}{2 v}$

$\Rightarrow x \frac{d v}{d x}=-\frac{\left(1+v^{2}\right)}{2 v}$

$\Rightarrow \frac{2 v}{1+v^{2}} d v=-\frac{d x}{x}$

Integrating both sides, we get:

$\log \left(1+v^{2}\right)=-\log x+\log \mathrm{C}=\log \frac{\mathrm{C}}{x}$

$\Rightarrow 1+v^{2}=\frac{\mathrm{C}}{x}$

$\Rightarrow\left[1+\frac{y^{2}}{x^{2}}\right]=\frac{\mathrm{C}}{x}$

$\Rightarrow x^{2}+y^{2}=\mathrm{C} x$

This is the required solution of the given differential equation.