Answer
$y^{\prime}=-\displaystyle \frac{y}{x(\ln x+1)}$
Work Step by Step
Differentiating an equation containing
terms of form $f(x)$ and $g(y)$,
terms $f(x)$ are differentiated directly, $\displaystyle \frac{d}{dx}[f(x)],$
and
terms $g(y)$ are differentiated using the chain rule,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
------------------
LHS: first term is a product, $(uv)^{\prime}=u^{\prime}v+uv^{\prime}$
RHS: constant
$y\cdot\ln x+y=2$
Differentiate both sides,
$(y^{\prime})\displaystyle \cdot\ln x+y\cdot(\frac{1}{x})+y^{\prime}=0\qquad/-\frac{y}{x}$
... isolating $y^{\prime}$ on one side...
$y^{\prime}\displaystyle \cdot\ln x+y^{\prime}=-\frac{y}{x}$
... on the LHS, factor out $y^{\prime}$
$y^{\prime}(\displaystyle \ln x+1)=-\frac{y}{x}\qquad/\div(\ln x+1)$
$y^{\prime}=-\displaystyle \frac{y}{x(\ln x+1)}$
