How does one prove that there exists a differentiable real-valued function $y(x)$ in some neighborhood of the point $x = 0$ such that $$x = e^{-y(x)}y(x)$$ for all $x$ in that neighborhood?
Furthermore, how would we show that this y(x) is a solution of the differentiable equation $xy' = y/(1-y)$?
Part 1
Consider the function $\mathrm{f}(x,y) = x-y\,\mathrm{e}^{-y}$. Your curve is the zero level-set $\mathrm{f}^{-1}(0) = \{(x,y): \mathrm{f}(x,y)=0\}$. The Jacobian Matrix of $\mathrm{f}$ is given by $$J_{\mathrm{f}}(x,y) = \left(1,(y-1)\mathrm{e}^{-y}\right)$$
For any value of $x$ and $y$, this matrix has maximal rank. It follows that there are no critical points of $\mathrm{f}$ and, more importantly, zero is a regular value. Hence $\mathrm{f}^{-1}(0)=\{(x,y) : x=y\,\mathrm{e}^{-y}\}$ is, in a neighbourhood of each of its points, a parametrisable one-dimensional manifold.
Note: the leading $1$ in $J_{\mathrm{f}}(x,y)$ tells is that $x$ can always be written as a function of $y$, but we knew that already! You're interested in a neighbourhood of $x=0$. Conisder $\mathrm{f}(0,y) = y\,\mathrm{e}^{-y}$. We want $\mathrm{f}(0,y)=0$ and hence $y=0$. When $y=0$ the Jacobian Matrix becomes $J_{\mathrm{f}}(0,0)=(1,-1)$.
Since the $y$-column is non-zero, (by the IFT), we can parametrise $\mathrm{f}^{-1}(0)$ as a function of $x$ in a sufficiently small neighbourhood of $(x,y)=(0,0)$.
Part 2
Consider the equation $x=y\,\mathrm{e}^{-y}$ where $y$ is a function of $x$. Differentiating with respect to $x$: $$x=y\,\mathrm{e}^{-y} \implies 1 = (1-y)\, y'\, \mathrm{e}^{-y} \implies y'=\frac{\mathrm{e}^{\,y}}{1-y} \implies xy'=\frac{y}{y-1}$$