Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a homogenous differential equation. Substitute $y=vx$
$\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{1}{2}[v-\frac{1}{v}]$
$\Rightarrow x\frac{dv}{dx}=-\frac{v^2+1}{2v}$
$\Rightarrow -\frac{2v}{v^2+1}dv=\frac{dx}{x}$
Integrating both sides
$\Rightarrow -\log|v^2+1|=\log x+\log c$
$\Rightarrow -\log|\frac{y^2}{x^2}+1|=\log xc$
$\Rightarrow -\log|\frac{x^2+y^2}{x^2}|=\log xc$
$\Rightarrow \frac{x^2}{x^2+y^2}= xc$
$\Rightarrow x= c(x^2+y^2)$
$\Rightarrow y=\pm \sqrt{xc-x^2}$
Kindly review my solution and let me know if there are other methods of solving such problems.

Your solution it is correct. Answering your second question, here is another slightly different solution.
If we can write the ODE as a Bernoulli equation: $y'+a(x)y=b(x)y^{\alpha}$ for some real $\alpha$ so the substitution $v=y^{1-\alpha}$ transforms the Bernoulli equation into a linear equation $\frac{1}{1-\alpha}v'+a(x)v=b(x)$ which can be solved using integrating factor.
We can re-write the ODE as
$$(x^2-y^2)dx+2xydy=0$$
$$(x^2-y^2)+2xy\frac{dy}{dx}=0$$
$$2xy\frac{dy}{dx}-y^2=-x^2$$
$$\frac{dy}{dx}-\frac{1}{2x}y=-\frac{x}{2}y^{-1}$$
Consider $a(x)=\frac{-1}{2x}$ and $b(x)=\frac{-x}{2}$ so the equation is Bernoulli and we can make the substitution $v=y^{1-(-1)}=y^{2}$ and then we can write the ODE as
$$\frac{1}{2}\frac{dv}{dx}-\frac{1}{2x}v=-\frac{x}{2}$$
$$\frac{dv}{dx}-\frac{1}{x}v=-x.$$
The equation it is linear and using your integrating factor $\mu(x)=e^{\int -\frac{1}{x}dx}=\frac{1}{x}$ we can transforms the linear equation into a separable equation.
$$\frac{d}{dx}\left(\mu(x)v\right)=-x\mu(x)$$
Integrating,
$$\frac{v}{x}=\int-1dx=-x+c$$
Thus,
$$v=-x^2+cx$$
Substitution back
$$y^2=-x^2+cx$$
Therefore, the general solution is given by
$$ y=\pm\sqrt{-x^2+cx}$$
as you said.
