Answer
$\frac{dy}{dx}=\frac{1-3y^2-2xy}{x^2+9xy^2}$
Work Step by Step
$x^2y + 3xy^3 − x = 3$
$\frac{d}{dx}( x^2y + 3xy^3 − x)=\frac{d}{dx}(3) $
$\frac{d}{dx}(x^2y)+3\frac{d}{dx}(xy^3)-\frac{dx}{dx}=0$
$x^2\frac{dy}{dx}+y\frac{d}{dx}(x^2)+3[x\frac{d}{dx}(y^3)+y^3\frac{dx}{dx}]-1=0 $
$x^2\frac{dy}{dx}+y(2x)+3[x\frac{d}{dy}(y^3 )\frac{dy}{dx}+y^3]-1=0 $
$x^2\frac{dy}{dx}+2xy+3[x(3y^2)\frac{dy}{dx}+y^3]-1=0 $
$x^2\frac{dy}{dx}+2xy+3[3xy^2\frac{dy}{dx}+y^3]-1=0 $
$x^2\frac{dy}{dx}+2xy+9xy^2\frac{dy}{dx}+3y^3-1=0 $
$x^2\frac{dy}{dx}+9xy^2\frac{dy}{dx}=1-3y^3-2xy $
$(x^2+9xy^2)\frac{dy}{dx}=1-3y^3-2xy $
$\frac{dy}{dx}=\frac{1-3y^2-2xy}{x^2+9xy^2}$
