Answer
(a) $e^{y} d y=-x^{-2} d x$
(b) $e^{y} =x^{-1}+C $
(c) $y =\ln \left(x^{-1}+C\right)$
(d) $y =\ln \left(x^{-1}+ e^4-0.5\right)$
Work Step by Step
Given
$$ x^{2} y'+e^{-y}=0$$
(a) Then
\begin{aligned} x^{2} \cdot \frac{d y}{d x} &=-e^{-y} \\ \frac{d y}{-e^{-y}} &=\frac{d x}{x^{2}} \\ e^{y} d y &=-x^{-2} d x \end{aligned}
(b) Integrate both sides
\begin{aligned} \int e^{y} d y &=\int-x^{-2} d x \\ e^{y} &=-\frac{x^{-1}}{(-1)}+C \\ e^{y} &=x^{-1}+C \end{aligned}
(c) Then
\begin{align*}
\ln \left(e^{y}\right)&=\ln \left(x^{-1}+C\right)\\
y&=\ln \left(x^{-1}+C\right)
\end{align*}
(d) At $x= 2 $, $y=4$, then $c= e^4-0.5$ and
$$y =\ln \left(x^{-1}+ e^4-0.5\right)$$
