Answer
${dy}=\frac{1-y}{(3\sqrt{y}+x)}{dx}$
Work Step by Step
We evaluate the function: $2\sqrt (y^3)+xy-x=0$
on differentiating the above:
$\frac{d{(2\sqrt (y^3)+xy-x})}{dx}=0$
or $2\times\frac{3}{2}\times\sqrt{y}\frac{dy}{dx}+x\frac{dy}{dx}+y-1=0$
or ${(3\sqrt{y}+x)}\frac{dy}{dx}=(1-y)$
or${dy}=\frac{1-y}{(3\sqrt{y}+x)}{dx}$
The final answer is:
${dy}=\frac{1-y}{(3\sqrt{y}+x)}{dx}$
