Let us restate the problem for ease of reference. The equations are
$\dot{x}(t)=y(t)$
$\dot{y}(t)= - 4 x(t) + y(t)^2$
Physically they describe an anharmonic oscillator with spatial coordinate $x(t)$ and velocity $y(t)$ in the potential
$U = 2 x^2 - \frac{1}{3} x^3$
and the energy
$E = \frac{1}{2} y^2 + U(x)$
The problem is then: given two solutions
$s_{1}(t) = (x_{1}(t), y_{1}(t))$
$s_{2}(t) = (x_{2}(t), y_{2}(t))$
corresponding to the initial conditions
$s_{1}(0) = (x_{1}(t=0) = x_{10}, y_{1}(t=0) = y_{10})$
$s_{2}(0) = (x_{2}(t=0) = x_{20}, y_{2}(t=0) = y_{20})$
Assuming that $y_{10}$ and $y_{20}$ do not vanish show that the quantity
$d(t) = y_{1}(t) + y_{2}(t)$
becomes $\neq 0$ for times $t>0$ even if it is $= 0$ for $t = 0$, i.e.
$y_{10} + y_{20} = 0$.
We prove it indirectly, assuming $d(t) = 0$ for all times.
The idea is to expand the solution $y(t)$ into a power series in $t$.
$y(t) = y(0) + t \dot{y}(0) + t^2 \frac{1}{2} \ddot{y}(0) + t^3 \frac{1}{6} \frac{\partial ^3y(0)}{\partial t^3} + ...$
In order to have $d(t) = 0$, all coefficients must vanish. These can be expressed through the initial values, and the first coefficients are
$c(0) = (y_{10} + y_{20}) $
$c(1) = - 4 (x_{10} + x_{20}) + (x_{10}^2 + x_{20}^2) $
$c(2) = - 4 (y_{10} + y_{20}) + 2 ( x_{10} y_{10} + x_{20} y_{20} ) $
$c(3) = 2 (y_{10}^2 + y_{20}^2) + 2 ( x_{10}^3+x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $
Since we have $y_{10} + y_{20} = 0$, $d(t) = 0$ requires
$c(1) = 0 = - 4 (x_{10} + x20) + (x_{10}^2 + x_{20}^2) $
$c(2) = 0 = 2 y_{10} (x_{10} - x_{20}) $
$c(3) = 0 = 4 y_{10}^2 + 2 ( x_{10}^3 + x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $
From $c(2) = 0$ and $y_{10} \neq 0$ we find $x_{20} = x_{10}$. Hence
$c(1) = 0 = - 4 x_{10} + x_{10}^2 $
$c(3) = 0 = y_{10}^2 + x_{10}(x_{10}-2)(x_{10}-4) $
Both solutions of $c(1) = 0$ give $y_{10} = 0$. The contradiction proves the assertion of the OP.
Observation: if we would drop the term $x^2$ in the second equation we can have $d(t) = 0$ by taking $x_{10} = x_{20}$