If $x=y\sqrt{1-y^2}$, then $\frac{dy}{dx}=$

$x=y\sqrt{1-y^2}$

Squaring on both sides

$x^2=y^2(1-y^2)$

$x^2=y^2-y^4$

Differentiate with respect to $x$

$2x=2y.y\text{'}-4y^3.y\text{'}$

$y\text{'}=\frac{2x}{2y-4y^3}$

$=\frac{x}{y-2y^3}$

$=\frac{x}{y(1-2y^2)}$

$=\frac{y\sqrt{1-y^2}}{y(1-2y^2)}$

$=\frac{\sqrt{1-y^2}}{1-2y^2}$