**Edited on May 2, 2020:** The OP pointed out that I had not addressed a special case (namely $C=1$ below), so I am amending my answer to address this and reorganizing so that the $C=1$ case gets addressed naturally when it comes up. — RLB

We can assume that $\phi$ is not constant, since the only constant solution is $\phi\equiv0$.

Thus, assume that the solution has $\phi'\not=0$ on some interval $I$. Multiply the equation $\phi^2 = \pm\phi''\sqrt{1-\phi'^2}$ by $\phi'$, then integrate both sides to get that there is a constant $C$ such that
$$
\phi^3 = C^3 \mp (1-\phi'^2)^{3/2}.
$$
The case $C=0$ corresponds to $\phi^2 = 1-\phi'^2$, which, since $\phi'$ is assumed nonzero on $I$ implies that $\phi = \cos(t-t_0)$ for some constant $t_0$. We can thus set $C=0$ aside and assume that $C\not=0$.

We have $C^3{-}1\le \phi^3\le C^3{+}1$, so $C\ge -1$, otherwise there cannot be any nonnegative solutions. If $C = -1$, then $\phi\le0$ and, since we are only interested in non-negative solutions, the only solution in this case is $\phi\equiv0$, so we can assume henceforth that $C>-1$.

Both equations (with either sign) can be studied as special cases of the polynomial differential equation
$$
(\phi^3-C^3)^2 - (1 - \phi'^2)^3 = 0,
$$
so that is what we will do. Set $\phi^3-C^3 = u^3$ where $|u|\le 1$
and note that this implies $(\phi')^2 = 1-u^2$. We then have
$$
\pm 1 = \frac{u^2u'}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}},
$$
so
$$
t_1-t_0 = \pm\int_{u(t_0)}^{u(t_1)}
\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}.
$$
For $C\not=0,1$ (remember that $C>-1$), the integral
$$
\int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}
= \int_{-1}^1\frac{u^2\,du}{(1-u^2)^{1/2}(C+u)^{2/3}(C^2-Cu+u^2)^{2/3}}
$$
converges. However, when $C=1$, the denominator contains a factor of $(1+u)^{1/2+2/3} = (1+u)^{7/6}$, so the integral diverges at $u=-1$.

In the case $C=1$, we can write down a parametrization of the graph $\bigl(t,\phi(t)\bigr)$ in the form $\bigl(t(v),\phi(t(v))\bigr)$ where
$|v|<\sqrt2$ with
$$
\phi(t(v)) = \bigl(1+(1-v^2)^3\bigr)^{1/3}
\quad\text{and}\quad
t(v) = \int_0^v \frac{2(1-s^2)^2\,ds}{(2-s^2)^{7/6}(1-s^2+s^4)^{2/3}}.
$$
Note that, while $t$ is a strictly increasing function of $v$ and $|t|\to\infty$ as $|v|\to\sqrt2$, $t'(\pm1) = 0$, and $\phi$ is not
a smooth function of $t$ where $v = \pm 1$. (It is, however, continuously once differentiable there, see below.)

Meanwhile, when $C\not=1$, a solution $\phi$ exists for all time and is periodic, as $\phi$ oscillates between $(C^3{-}1)^{1/3}$ and $(C^3{+}1)^{1/3}$. The period of $\phi$ is
$$
p(C) = 2\int_{(C^3{-}1)^{1/3}}^{(C^3{+}1)^{1/3}}\frac{d\xi}{\sqrt{1-(\xi^3-C^3)^{2/3}}}
= \int_{-1}^1\frac{2u^2\,du}{(1-u^2)^{1/2}(C^3+u^3)^{2/3}}.
$$
Of course, $p(0) = 2\pi$ and $p(C)$ has a series expansion $2\pi\,C^{-2}+\tfrac{175}{576}\pi\,C^{-8}+\cdots$ when $C>1$.

Near a minimum at $t=t_0$, the solution has a series expansion of the form
$$
\phi(t) = (C^3{-}1)^{1/3}\left(1+\tfrac12(C^3{-}1)^{1/3}(t-t_0)^2+\tfrac1{24}(C^3+1)(C^3{-}1)^{2/3}(t-t_0)^4 + \cdots\right)
$$
Near a maximum at $t=t_1$ the solution has a series expansion of the form
$$
\phi(t) = (C^3{+}1)^{1/3}\left(1-\tfrac12(C^3{+}1)^{1/3}(t-t_1)^2-\tfrac1{24}(C^3-1)(C^3{+}1)^{2/3}(t-t_1)^4 + \cdots\right)
$$

Note that, when $C\not=0$, $\phi$ is not $C^2$ when it attains the intermediate value $C$. In fact, if $\phi(t_2) = C$
and $\phi'(t_2) = 1$, then $\phi$ has a series expansion in powers of $(t{-}t_2)^{1/3}$:
$$
\phi(t) = C +(t{-}t_2) - \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3}
- \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots.
$$
Meanwhile, if $\phi(t_3) = C$
and $\phi'(t_3) = -1$, then $\phi$ has a series expansion in powers of $(t{-}t_3)^{1/3}$:
$$
\phi(t) = C -(t{-}t_2) + \frac{C^{4/3}}{10}\,\bigl(3(t{-}t_2)\bigr)^{5/3}
+ \frac{C^{8/3}}{280}\,\bigl(3(t{-}t_2)\bigr)^{7/3} + \cdots.
$$
Finally, note that $\phi''<0$ when $C<\phi<(C^3+1)^{1/3}$ while $\phi''>0$ when $(C^3-1)^{1/3}<\phi<C$.

individually? Or are you asking, like @DanieleTampieri suggested, for a $\phi$ that satisfiesat least oneof the two equations? If $\phi$ is allowed to follow exactly one (but not the other) equation, then by convexity there cannot be any global in time solutions. $\endgroup$ – Willie Wong Apr 28 at 12:01