quadratic approximation of $e^{x^3}$ at $x=0$ we know that
$$P(x)=f(a)+f'(a)x+\frac{f''(a)}{2}x^2 $$
therefore
$f(0)=1$
if $x^3=u$, so $$f'(e^u)=e^{x^3}3x^2$$ and $$f''(e^u) = 4(e^u)'3x^2+e^u (3x^2)'=e^{x^3}3x^23x^2+e^{x^3}6x=e^{x^3}(9x^4+6x)$$
$$P(x)=1+e^{x^3}3x^3+\frac{e^{x^3}(9x^4+6x)}{2}x^2 $$
is it correct?
can you please point out my mistake.
cause the system marked my answer as a wrong one.  
and what is the magnitude of the error?
 A: The quadratic approximation is just $1$. 
When you plug $a=0$ into the derivative you computed you get $0$, same for the second derivative. 
A: If you want a quadratic approximation then the alarm bells ought to start ringing. You have $P(x) = 1 + 3x^3\mathrm{e}^{x^3} + \cdots$ which is not a quadratic. It is not even a polynomial!
Recall the definition of the Maclaurin Series:
$$\mathrm{f}(x) \sim \mathrm{f}(0) + \mathrm{f}'(0)x + \frac{\mathrm{f}''(0)}{2!}x^2 + \cdots $$
where $\mathrm{f}(0)$, $\mathrm{f}'(0)$ and $\mathrm{f}''(0)$ are numbers.
First, you need to differentiate $\mathrm{f}(x) = \mathrm{e}^{x^3}$ once to get $\mathrm{f}'(x)$ and then again to get $\mathrm{f}''(x)$. 
Second, substitute $x=0$ into $\mathrm{f}(x)$, $\mathrm{f}'(x)$ and $\mathrm{f}''(x)$ to get $\mathrm{f}(0)$, $\mathrm{f}'(0)$, and $\mathrm{f}''(0)$. 
Third, put your values for $\mathrm{f}(0)$, $\mathrm{f}'(0)$, and $\mathrm{f}''(0)$ into the general formula 
$$\mathrm{f}(x) \sim \mathrm{f}(0) + \mathrm{f}'(0)x + \frac{\mathrm{f}''(0)}{2!}x^2 + \cdots $$
Let me get you started: $\mathrm{f}(x) = \mathrm{e}^{x^3}$ and so $\mathrm{f}'(x) = 3x^2\mathrm{e}^{x^3}$. Putting $x=0$ gives $\mathrm{f}'(0) = 3 \times 0^2 \times \mathrm{e}^{0^3} = 0$. You need to find the numbers $\mathrm{f}(0)$ and $\mathrm{f}''(0)$.
A: We can just kick it and do: $$e^x = \sum_{n \geq 0}\frac{x^n}{n!} \implies e^{x^3} = \sum_{n \geq 0}\frac{x^{3n}}{n!},$$so that: $$e^{x^3} = 1 + x^3 + \frac{x^6}{2} + \frac{x^9}{6}+\cdots.$$We have no $x$ and $x^2$ terms.
