For simplicity, multiply the equation by $2x$ and subtract $2t$ to get $ (x^2-t^2)'=2x^2(x^2-t^2)-2t\tag{1} $ First, let's deal with $t>0$.
$(1)$ says that if $x^2-t^2<0$, then we also have $(x^2-t^2)'<0$. Therefore, if $x(t_0)^2-t_0^2<0$, we have $x(t)^2-t^2<0$ for all $t\ge t_0$.
Next, let's deal with $t<0$.
$(1)$ says that if $x^2-t^2\ge0$, then we also have $(x^2-t^2)'\ge0$. Therefore, if $x(t)^2-t^2\ge0$, we have $x(t_0)^2-t_0^2\ge0$ for all $t_0\ge t$. The contrapositive says that for all $t\le t_0$, if $x(t_0)^2-t_0^2<0$, then we have $x(t)^2-t^2<0$.
A Graphical Explanation
Consider the direction field (aka slope field and flow diagram) for the solutions. That is, the vector field $\color{#0000c0}{\frac{\mathrm{d}}{\mathrm{d}t}(t,x)}$ which is tangent to all solutions $\color{#c00000}{(t,x)}$.
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The important part to consider is the flow across the boundary $x^2=t^2$ between the regions where $\color{#00c000}{x^2< t^2}$ and $\color{#c000c0}{x^2>t^2}$. Notice that the on the boundary, the flow is horizontal since $x'=x(x^2-t^2)=0$.
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Thus, when $t<0$ (on the left), a solution can only move from the region where $\color{#00c000}{x^2< t^2}$ to the region where $\color{#c000c0}{x^2>t^2}$. When $t>0$ (on the right), a solution can only move from the region where $\color{#c000c0}{x^2>t^2}$ to the region where $\color{#00c000}{x^2< t^2}$.