$\displaystyle
y - x\frac{dy}{dx} = \sqrt{1 + (\frac{dy}{dx})^2}(x^2 + y^2)
$
I Have Tried It At my level best but was unable to solve it!
$\displaystyle y - x\,\frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,(x^2 + y^2)$
$\displaystyle \left(y - x\,\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^2 + y^2)^2$
$\displaystyle y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = \left[1 + \left(\frac{dy}{dx}\right)^2\right](x^4 + 2x^2y^2 + y^4)$
$\displaystyle y^2 - 2xy\,\frac{dy}{dx} + \left(\frac{dy}{dx}\right)^2 = x^4 + 2x^2y^2 + y^4 + (x^4 + 2x^2y^2 + y^4)\left(\frac{dy}{dx}\right)^2$
$\displaystyle (x^4 + 2x^2y^2 + y^4 - 1)\left(\frac{dy}{dx}\right)^2 + 2xy\,\frac{dy}{dx} + x^4 + 2x^2y^2 + y^4 - y^2 = 0$.
This is now a Quadratic in $\displaystyle \frac{dy}{dx}$, so solve using the Quadratic Formula.
Just use the Quadratic formula. It's an equation of the form
$\displaystyle aX^2 + bX + c = 0$.
Here $\displaystyle X = \frac{dy}{dx}, a = x^4 + 2x^2y^2 + y^4 -1, b = 2xy, c = x^4 + 2x^2y^2 + y^4 - y^2$.
So the solutions are
$\displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
It's ok to keep your solution in terms of $\displaystyle x$ and $\displaystyle y$. You don't need to do any integration at all.