Solve this

We are given with an equation $x y \log (x+y)=1$, we have to prove that $\frac{d y}{d x}=\frac{y\left(x^{2} y+x+y\right)}{x\left(y^{2} x+x+y\right)}$ 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 the triple product rule, which is, $\frac{\mathrm{d}(\mathrm{uvw})}{\mathrm{dx}}=\mathrm{uw} \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{vw} \frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{uv} \frac{\mathrm{dw}}{\mathrm{dx}}$,

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

From the equation put $\log (x+y)=\frac{1}{x y}$

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

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

$\frac{x^{2} y+x+y}{(x+y) x}+\frac{d y}{d x}\left[\frac{y^{2} x+x+y}{(x+y) y}\right]=0$

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