Find interval with function that solves ODE $y'(x)=1+(y(x))^2$ Let $g\in C^1(\mathbb{R})$ with $g'\gt 0$ and $g(0)=0$.
Show that for the differential equation $$\begin{cases}y'(x) & = \dfrac{1}{g'(y(x))} \\[8pt] y(0) & = 0  \\\end{cases}$$ there exists exactly one non-empty open interval $I\subseteq\mathbb{R}$ and an $f\in C^1(I)$, such that $f$ solves the differential equation, $0\in I$ and $(I,f)$ is a maximum solution.

Hint: as $g'\neq 0$, using the chain rule the equation is equivalent to 
$$
g(y(x)) = g(y(0)),
\\
y(0) = 0
$$
