= Solution
Use the <product substitution for a first-order equation> $v=xy$. The <product rule> gives $v'=xy'+y$, turning the equation into $v'=e^v$. Consequently $(e^{-v})'=-1$, so $e^{-v}=C-x$.
At $x=1$ the initial value gives $v=1$, hence $C=1+e^{-1}$. Recovering $y=v/x$ gives
$$
\boxed{y(x)=-\frac{\ln(1+e^{-1}-x)}{x},\qquad 0<x<1+e^{-1}.}
$$
This is the maximal real graph <interval> containing the initial point: the logarithm requires $x<1+e^{-1}$, and the numerator has a nonzero limit at zero, so this <particular solution> cannot extend through zero. Differentiating $xy=-\ln(C-x)$ verifies $xy'+y=1/(C-x)=e^{xy}$.
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