Solve the ODE $y'' = (y')^2$ I am asked in a past question paper to solve the following ODE: $$ y'' = (y')^{2}$$
To solve this, I began by equating $$y' = u$$
Differentiating both sides $w.r.t.x$ $$ \frac{d^{2}y}{dx^{2}} = \frac{du}{dx}$$
And therefore substituting back to out original equation I got:
$$  \frac{du}{dx} = u^{2}$$
Rearranging $$\frac{du}{u^{2}} = dx$$
Integrating $w.r.t.x$ I got
$$ \frac{-1}{u} = x+C$$ and substituting back $u$
$$ \frac{-dx}{dy} = x+C$$
$$ =x + \frac{dx}{dy} = C$$
$$=\frac{-dx}{x+c} = dy$$
$$ \log(\frac{1}{x+c}) = y$$
$$ -(x+c) = e^{y}$$
Is my answer and method correct?
EDIT As pointed out by @MtGlasser
We can re-write the D.E as $$ \frac{y''}{y'} = y'$$
Which is of the form :
$$(ln(y'))' = y'$$
Integrating on both sides we get
$$ ln (y') = y + c$$
And taking exponent on both sides and re-arranging we get
$$ e^{-1}e^{-y} dy = dx$$
Integrating we get the asnwer $$ e^{-(y+1)} - C = x$$

Your method is mostly correct, although the steps could be made more clear for the latter part and also you didn't include another constant of integrating after the last integration. Note you can stop once you have expressed $y$ in terms of $x$, unless it says in the question to express $x$ in terms of $y$.
We have
$$-\frac{dx}{dy}=x+C_{1}$$
$$\frac{dx}{dy}=-x-C_{1}$$
$$-\frac{dx}{x+C_{1}}=dy$$
$$\color{red}{C_{2}}-\ln(x+C_{1})=y$$
$$\implies y(x)=C_{2}-\ln(x+C_{1})$$
