$u = x^2 \implies du = 2x dx$

$v = y^2 \implies dv = 2y dy$

$p' = \frac{dv}{du} = \frac{y}{x}\frac{dy}{dx} = \frac{y}{x}p \\ \implies p = \frac{x}{y}p'$

Substituing in the given differential equation,

$xyp'^2\frac{x^2}{y^2} - (x^2+y^2-1)p'\frac{x}{y} + xy = 0 \\ \implies x^2p'^2-(x^2+y^2-1)p' + y^2 = 0 \\ \implies up'^2 - (u+v-1)p'+v=0 \\ \implies (p'-1)v = up'(p'-1)-p' \\ \implies v = up' - \frac{p'}{p'-1}$

To solve the given differential equation, just replace $p'$ by $c$ in the Clairaut's equation

$\implies v = uc-\frac{c}{c-1} \implies y^2 = cx^2 - \frac{c}{c-1}$