Solve this

Given $x y=4$

$\Rightarrow y=\frac{4}{x}$

On differentiating y with respect to $x$, we get

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{4}{\mathrm{x}}\right)$\

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4 \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{1}{\mathrm{x}}\right)$

$\Rightarrow \frac{d y}{d x}=4 \frac{d}{d x}\left(x^{-1}\right)$

We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=4\left(-1 \mathrm{x}^{-1-1}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-4 \mathrm{x}^{-2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{4}{\mathrm{x}^{2}}$

Now, we will evaluate the LHS of the given equation.

$x\left(\frac{d y}{d x}+y^{2}\right)=x\left(-\frac{4}{x^{2}}+y^{2}\right)$

$\Rightarrow x\left(\frac{d y}{d x}+y^{2}\right)=x\left(\frac{-4+x^{2} y^{2}}{x^{2}}\right)$

$\Rightarrow x\left(\frac{d y}{d x}+y^{2}\right)=\frac{x^{2} y^{2}-4}{x}$

$\Rightarrow x\left(\frac{d y}{d x}+y^{2}\right)=\frac{(x y)^{2}-4}{x}$

However, $x y=4$

$\Rightarrow \mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2}\right)=\frac{12}{\mathrm{x}}$

$\Rightarrow \mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2}\right)=3\left(\frac{4}{\mathrm{x}}\right)$

$\therefore \mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2}\right)=3 \mathrm{y}[\because \mathrm{xy}=4]$

Thus, $\mathrm{x}\left(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2}\right)=3 \mathrm{y}$