Let $f:\mathbb{R}\rightarrow\mathbb{R}$ a continuous function and $\phi$ a solution on its maximal interval of existence for the initial value problem $$ \dot{x}=f(t)x^2\hspace{0.3cm},\hspace{0.3cm}x(0)=1 $$ Show that the solution exists on $\mathbb{R}$ iff $$ \int_{0}^{t}f(s)ds<1\hspace{0.3cm}\forall t\in\mathbb{R} $$
So what I think would be the right way to start is to note that $\dfrac{\dot{x}}{x^2}=-\dfrac{d}{dt}\dfrac{1}{x}$ however that only works when $x\neq 0$ so I'm not quite sure about how to proceed.
The right-hand side $$ F(t, x) = f(t) x^2 $$ is locally Lipschitz continuous in $x$, so that every solution of an initial value problem $$ \dot x = F(t, x(t)) \, , \quad x(t_0) = x_0 $$ on an interval is unique. It follows that for every solution on an interval, either $x(t) \equiv 0$ or $x(t)$ has no zeros.
In your case $x(0) = 1$ so that $x(t) \ne 0$ (actually: $x(t) > 0$) on the maximal interval of existence, and you can proceed by integrating $\frac{\dot{x}}{x^2}$ (i.e. by a “separation of variables”): If a solution on $\Bbb R$ exists then $$ \tag{*} \int_{0}^{t} f(s) \,ds = \int_{0}^{t}\frac{\dot x(s)}{x^2(s)} \, ds = 1 - \frac{1}{x(t)} $$ is less than one.
And if $\int_{0}^{t} f(s) \,ds < 1$ for all $t$ then $(*)$ can be solved for $x(t)$ to get a solution on $\Bbb R$.