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As in title guys, I need to solve it analytically, but I really don't have a clue what method to use. I don't expect you to solve it for me, just direct me if possible. Thanks.

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    I don't know much about differential equations but it seems to me you could divide by $y'$ on both sides to obtain $$\frac{y''}{y'} = -\frac{y'}{y}$$ and then integrate. – 2017-01-31
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    Have you ever seen the reverse product rule in action? This seems like the kind of diff. equations where you're supposed to learn how to use it. – 2017-01-31
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    $$0=y''y+(y')^2=(y'y)'=\tfrac12(y^2)''\implies y^2=ax+b$$ – 2017-01-31

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$yy''=-y'^{2} \Rightarrow (y'y)'=0 \Rightarrow yy'=c \Rightarrow y^2/2=\int yy'=cx+d$

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    Worth noting for general knowledge, that this approach is usually referred to as the "reverse product rule", for obvious reasons. – 2017-01-31
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I suggested in a comment that it might be fruitful to divide by $y'$ on both sides to obtain $$\frac{y''}{y'} = -\frac{y'}{y}$$ and then integrate. Following through on this we get $$\begin{align} \ln y' & = -\ln y +\ldots \\ \ln yy'& = \ldots \\ yy' = \ldots \end{align}$$

Where each of the $\ldots$ represents some constant of integration.

Since it's basic calculus that $yy' = \frac12(y^2)'$ we get $$ y = \pm\sqrt{C_1x+C_0}$$

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    The third step in your calculation is misleading. It's true that you can follow from $\ln (yy') = C$ that $yy'$ is a constant but you cannot and should not use the same $C$ to denote that constant, since it will actually be $e^C$. Just introduce a new name $C_2$ or $\alpha$ or whatever to be completely consistent in your notation. – 2017-01-31
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    I have replaced the $C$ with $\ldots$ so as to avoid the suggestion that the various constants are equal. – 2017-01-31
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$$y''=-\frac{(y')^2}{y}$$ in such differential equations without $x$, simply could replace $y''=uu'$, $y'=u$ and $y$ is independent variable, so $$uu'=-\frac{u^2}{y}$$ this concludes that $u=0$ or $\dfrac{du}{u}=-\dfrac{1}{y}dy$ then $y'=0$ or $\ln y'=-\ln y+C$ that is $y'=\dfrac{C}{y}$. Solutions are $y=C$ or $y^2=Cx+C'$.

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\begin{align*} y' &= \frac{dy}{dx} \\ y'' &= \frac{dy'}{dx} \\ &= \frac{dy'}{dy} \times \frac{dy}{dx} \\ &= y' \frac{dy'}{dy} \\ y' \frac{dy'}{dy} &= -\frac{(y')^2}{y} \\ \end{align*}

If $y'=0$, then $$y=C$$

If $y' \ne 0$, then \begin{align*} \frac{dy'}{y'} &= -\frac{dy}{y} \\ \ln y' &= -\ln y+C \\ y' &= \frac{A}{y} \\ \frac{dy}{dx} &= \frac{A}{y} \\ \int y\, dy &= \int A \, dx \\ y^2 &= Ax+B \\ y &= \pm \sqrt{Ax+B} \end{align*}