Prove this solution is genuine I cannot prove that the solution to :$$y'=-\frac{x\sqrt{1-y^2}}{y\sqrt{1-x^2}}$$
is:
$$y=\sqrt{x^2-2C\sqrt{-x^2+1}-C^2},\:y=-\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}$$
By substituting the solution(s) back into the original equation. I have taken derivative and attempted to equate LHS with RHS but I just can't simplify enough to obviously justify the solution(s) $$\frac{x\left(C+\sqrt{-x^2+1}\right)}{\sqrt{-x^2+1}\sqrt{x^2-2C\sqrt{-x^2+1}-C^2}}$$

Square the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$
Taking the derivative:
$$2yy'= 2x+2C\frac{x}{\sqrt{1-x^2}}$$
So
$$yy' = x + C\frac{x}{\sqrt{1-x^2}}=\frac{x\sqrt{1-y^2}}{\sqrt{1-x^2}}$$
Assuming $x$ is not $0$, then
$$\sqrt{1-x^2} + C=\sqrt{1-y^2}$$
Squaring both sides:
$$1-x^2+2C\sqrt{1-x^2}+C^2=1-y^2$$
Then,
$$y^2=x^2-2C\sqrt{1-x^2}-C^2$$
QED
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The solution to this ODE is by first noticing it is separable. Then you arrive at the equation:
$$-\sqrt{1-y^2}=\sqrt{1-x^2}+C$$
This gives you the solution:
$$y^2 = x^2-2C\sqrt{1-x^2}-C^2$$
