Answer
$x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+k$
Work Step by Step
We are given:
$y''+y^{-1}(y')^2=ye^{-y}(y')^3$
Suppose that
$u=\frac{dy}{dx}$
Now, $\frac{du}{dx}=\frac{d^2y}{dx^2}$
Subsitute to the equation:
$u\frac{du}{dx}+\frac{u^2}{y}=ye^{-y}u^3$
Multiplying both sides by $\frac{1}{u}$
$\frac{du}{dy}+\frac{u}{y}=ye^{-y}u^2$
Then continue to multiply both sides by $\frac{1}{u^2}$:
$\frac{1}{u^2}\frac{du}{dy}+\frac{1}{uy}=ye^{-y}$
Let $t=\frac{1}{u} \rightarrow \frac{-du}{u^2}=dt$
$-\frac{dt}{dy}+\frac{t}{y}=ye^{-y}$
$\frac{dt}{dy}-\frac{1}{y}t=ye^{-y}$
Integrating factor:
$I=e^{-\int \frac{1}{y}dy}=e^{-\ln y}=y^{-1}$
Integrating both sides:
$\frac{1}{y}t=e^{-y}+C$
$\frac{1}{u}=ye^{-y}+Cy$
$u=\frac{1}{ye^{-y}+Cy}$
$\frac{dy}{dx}=\frac{1}{ye^{-y}+Cy}$
Integrating the last equation:
$x=\int ye^{-y}dy + C\int ydy$
$x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+C_2$
The final solution is:
$x=-ye^{-y}-e^{-y}+C\frac{y^2}{2}+k$