Solve this

We are given with an equation $x y^{2}=1$, we have to prove that $2 \frac{d y}{d x}+y^{3}=0$ by using the given equation we

will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

$y^{2}(1)+2 x y \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{-y}{2 x}$

Or we can further solve it by using the given equation,

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}^{3}}{2}$

By putting this value in the L.H.S. of the equation, we get,

$2\left(\frac{-y^{2}}{2}\right)+y^{3}=0=$ R.H.S.