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Let $f$ be thrice differentiable function on $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ such that $f'(x)= 1+(f(x))^2$.

If $f(0)=1$, then the coefficient of $x^2$ in Taylor's expansion of $f$ about $0$ is?

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    for a polynomial the expansion is the polynomial itself. Thus the coefficient is 1. – 2017-02-09
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You know that $$ f''(x)=2f'(x)f(x) $$ by the chain rule. Also, $f'(0)=1+(f(0))^2=2$, so $$ f''(0)=2\cdot2\cdot1=4 $$

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    On the other hand, the solution of the Cauchy problem $y'=1+y^2$, $y(0)=1$ cannot be a function defined over $[-\pi/2,\pi/2]$. – 2017-02-09
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    what do you mean by this? The solution is clearly $\tan(x+\pi / 4)$... Are you referring to the discontinuity at $\frac{\pi}{4}$? – 2017-02-09
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    @BrevanEllefsen Right so. Is there a solution that is differentiable over $[-\pi/2,\pi/2]$? I think not. – 2017-02-09