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

It looks alright.
You're mistaken in saying there's no ODE. Every line in your posting is just an ODE.
From where you left off I'd go on to say
$$
\int\pm\sqrt{\frac{-x^2}{x^2-1}}\,dx = \int\pm\sqrt{\frac{x^2}{1-x^2}} = \int\pm \frac x {\sqrt{1-x^2}} \, dx = \int\pm \frac {du}{2\sqrt{u}} = \pm \sqrt u + C
$$
$$
= \pm \sqrt{1-x^2} + C.
$$
