How to solve this ODE: $y'(x) e^x = y^2(x)$?

As mentioned in the comments, Wolfram Alpha interprets $y^2(x)$ as $y^2times x$. You can write $(y(x)^2)$ in Wolfram Alpha instead. The correct input in Wolfram Alpha produces:

$$y(x)=-frac{e^x}{Ce^x-1}$$ In your solution, $b(x)=e^{-x}$. Setting $u(x) = y^{-1} implies u'(x) = -y^{-2}(x) y'(x)$, so you should divide both sides of the original differential equation by $y^2$

$$y'(x) e^x = y^2(x)implies frac{y'(x)}{y^2(x)}=e^{-x}implies -u'(x)=e^{-x}$$

Hence

$$-frac{du(x)}{dx}=e^{-x}$$ $$-u(x)=-e^{-x}+C$$ $$-frac{1}{y(x)}=-e^{-x}+C$$ $$y(x)=-frac{1}{C-e^{-x}}$$

Therefore, multiplying the numerator and denominator by $e^x$ forms $$y(x)=-frac{e^{x}}{Ce^{x}-1}$$

Alternatively, we may write

$$ y(x) = frac{1}{e^{-x}+C}$$

Note: $y(x)=0$ is not a solution.

Your equation is equivalent to $$frac{dot{y}}{y^2}=e^{-x}$$ as long as $y(t)neq0$. (Notice that $y(t)equiv0$ is a solution to your problem)

Integrating over some intervals,say $[x_0,x]$ leads to

$$ int^x_{x_0}frac{y'(t)}{y^2(t)},dt=int^x_{x_0}e^{-t},dt=-e^{-t}|^x_{x_0}=e^{-x_0}-e^{-x} $$

The integral on the left can be simplifies by change of variables $u=y$ to get

$$ -frac{1}{y(t)}Big|^x_{x_0}=frac{1}{y(x_0)}-frac{1}{y(x)}=e^{-x_0}-e^{-x} $$

solving for $y(x)$ on gets

$$ frac{1}{y(x)}=frac{1}{y(x_0)}-e^{-x_0}+e^{-x} $$ and so $$ y(x)=frac{1}{y(x_0)^{-1}-e^{-x_0}+e^{-x}} $$

$$dfrac{dy}{dx}e^x=y^2$$ $$dfrac{dy}{y^2}=e^{-x}dx$$ $$int y^{-2}dy=int e^{-x}dx$$ $$-y^{-1}=-e^{-x}+c_1$$ $${1over y}=e^{-x}+c_2$$