Implicit differentiation involving arcsin

Differentiate both sides with respect to $x$. The left side gives $dy/dx$. For the right side, apply the chain rule to $\arcsin(xy)$: $\frac{d}{dx}\big[\arcsin(xy)\big] = \frac{1}{\sqrt{1-(xy)^2}}\cdot \frac{d}{dx}(xy).$ By the product rule, $\frac{d}{dx}(xy) = y + x\frac{dy}{dx}$, so the equation becomes $\frac{dy}{dx} = \frac{y + x\frac{dy}{dx}}{\sqrt{1-x^2y^2}}.$

Multiply both sides by $\sqrt{1-x^2y^2}$ to clear the denominator: $\frac{dy}{dx}\sqrt{1-x^2y^2} = y + x\frac{dy}{dx}.$ Collect all terms with $dy/dx$ on the left: $\frac{dy}{dx}\left(\sqrt{1-x^2y^2} - x\right) = y.$ Solving for $dy/dx$ gives the final result $\frac{dy}{dx} = \frac{y}{\sqrt{1-x^2y^2} - x}.$