Solve this

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

By using product rule, we get,

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

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y}}{\mathrm{x}}$

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

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

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

$-y^{2}+y^{2}=0=$ R.H.S.