If $x y \log (x+y)=1$, prove that $\frac{d y}{d x}=\frac{y\left(x^{2} y+x+y\right)}{x\left(x y^{2}+x+y\right)}$
Here,
$x y \log (x+y)=1 \ldots \ldots(i)$
Differentiating it with respect to $x$ using the chain rule and product rule,
$\frac{d y}{d x}(x y \log (x+y))=\frac{d}{d x}(1)$
$x y \frac{d}{d x} \log (x+y)+x \log (x+y) \frac{d y}{d x}+y \log (x+y) \frac{d}{d x}(x)=0$
$\left(\frac{x y}{(x+y)}\right)(1+)+x \log (x+y) \frac{d y}{d x}+y \log (x+y)=0$
$\left(\frac{x y}{(x+y)}\right) \frac{d y}{d x}+\frac{x y}{(x+y)}+x\left(\frac{1}{x y}\right) \frac{d y}{d x}+y\left(\frac{1}{x y}\right)=0$ [Using (i)]
$\frac{d y}{d x}\left[\frac{x y}{(x+y)}+\frac{1}{y}\right]=-\left[\frac{x y}{(x+y)}+\frac{1}{x}\right]$
$\frac{d y}{d x}\left[\frac{x y^{2}+x+y}{(x+y) y}\right]=-\left[\frac{x^{2} y+x+y}{(x+y) x}\right]$
$\frac{d y}{d x}=-\left[\frac{x^{2} y+x+y}{(x+y) x}\right] \times\left[\frac{(x+y) x}{x y^{2}+x+y}\right]$
$\frac{d y}{d x}=-\frac{y}{x}\left(\frac{x^{2} y+x+y}{x y^{2}+x+y}\right)$
Hence Proved.