We have the functional $I(x)=\int^3_2 y^2(1-y')^2dx$ and we want to find an extremal that satisfies some boundary conditions.
I found the Euler equation (I think):
\begin{align*} \frac{\partial F}{\partial y'} &= -2y^2(1 - y') \implies \frac{d}{dx}\bigg(\frac{\partial F}{\partial y'}\bigg) = -4yy'(1-y') +2y^2y'' \\ \frac{\partial F}{\partial y} &= 2y(1-y')^2 \end{align*}
We get the equation $2y(1-y')^2 = -4yy'(1-y') +2y^2y''$:
\begin{align*} &2(1-y')^2 = -4y'(1-y') +2yy''\\ \iff& 2(1-2y' + y'^2) = -4y'(1-y') +2yy''\\ \iff& 2-4y' + 2y'^2 = -4y' +4y'^2 +2yy''\\ \iff& 1 - y'^2 = yy''\\ \end{align*}
I don;t know how to solve that, so by inspection I am really tempted to say it must be a trig function because if the equation was $-yy'' = 1-y'^2$ $\cos$ would totally be a solution.