Answer
$y^{\prime}=\displaystyle \frac{e^{x}y}{1-e^{x}}$
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,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
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LHS: first term is a product, $(uv)^{\prime}=u^{\prime}v+uv^{\prime}$
RHS: constant
$e^{x}\cdot y-y=2$
Differentiate both sides,
$(e^{x})\cdot y+e^{x}(y^{\prime})-y^{\prime}=0\qquad/-e^{x}y$
... isolating $y^{\prime}$ on one side...
$e^{x}(y^{\prime})-y^{\prime}=-e^{x}y$
... on the LHS, factor out $y^{\prime}$
$y^{\prime}(e^{x}-1)=-e^{x}y\qquad/\div(e^{x}-1)$
$y^{\prime}=\displaystyle \frac{-e^{x}y}{(e^{x}-1)}=\frac{-e^{x}y}{-(1-e^{x})}$
$y^{\prime}=\displaystyle \frac{e^{x}y}{1-e^{x}}$
