<s>[INST] 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
 [/INST] 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
$$
\int\pm\sqrt{\frac{-x^2}{x^2-1}}\,dx = \int\pm\sqrt{\frac{x^2}{1-x^2}} = \int\pm \frac x {\sqrt{1-x^2}} \, dx = \int\pm \frac {du}{2\sqrt{u}} = \pm \sqrt u + C
$$
$$
= \pm \sqrt{1-x^2} + C.
$$
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