Solve $x=\pm\frac{y'}{\sqrt{(y')^2+1}}$ 
$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$

$$x=\pm\frac{y'}{\sqrt{(y')^2+1}}$$
$$x^2=\frac{(y')^2}{{(y')^2+1}}$$
$$x^2(y')^2+x^2=(y')^2$$
$$(y')^2[x^2-1]=-x^2$$
$$(y')^2=\frac{-x^2}{x^2-1}$$
$$y'=\pm \sqrt{\frac{-x^2}{x^2-1}}$$
Have I got it wrong? as there is no ODE

Here is a solution

$$y'=\pm \sqrt {\frac{-x^2}{x^2-1}} \\ \frac{dy}{dx} =\pm \frac x {\sqrt {1-x^2}} \\ \int dy = \pm \int \frac x {\sqrt {1-x^2}} \, dx =\pm \frac 1 2 \int \frac {d(1-x^2)}{\sqrt{ 1-x^2}}=\pm \sqrt {1-x^2} \\ y=\pm \sqrt{1-x^2} +C$$

