By definition, the Maclaurin series for [tex]f(x)[/tex] is
[tex]\displaystyle\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}[/tex]
It sounds to me like you're asking why [tex]f^{(n)}(0)[/tex] is zero for some terms, and non-zero for others. Take the derivative to see why this is the case.
[tex]f(x)=f^{(0)}(x)=x^2e^{x^2}[/tex]
[tex]f^{(1)}(x)=2xe^{x^2}+x^2(2x)e^{x^2}=(2x+2x^3)e^{x^2}[/tex]
[tex]f^{(2)}(x)=(2+6x^2)e^{x^2}+(2x+2x^3)(2x)e^{x^2}=(2+10x^2+4x^4)e^{x^2}[/tex]
[tex]f^{(3)}(x)=(20x+16x^3)e^{x^2}+(2+10x^2+4x^4)(2x)e^{x^2}=(24x+36x^3+8x^4)e^{x^2}[/tex]
and so on. There's a pattern here. Every odd-order derivative involves an odd-degree polynomial multiplied by [tex]e^{x^2}[/tex] with no constant term. So [tex]f^{(n)}(0)=0[/tex] when [tex]n[/tex] is odd. Meanwhile, every even-order derivative consists of an even-degree polynomial multiplied by [tex]e^{x^2}[/tex] and *does* have a constant term, which is why [tex]f^{(n)}(0)\neq0[/tex] when [tex]n[/tex] is even.
This would explain the absence of odd powers of [tex]x[/tex] in the series expansion for [tex]f(x)[/tex].
But you can see why this must happen just by referring to the manipulated series for [tex]e^x[/tex]:
[tex]e^x=\displaystyle\sum_{n=0}^\infty\frac{x^n}{n!}=x^0+x^1+\frac{x^2}2+\frac{x^3}6+\cdots[/tex]
(both even and odd powers)
[tex]e^{x^2}=\displaystyle\sum_{n=0}^\infty\frac{x^{2n}}{n!}=x^0+x^2+\frac{x^4}2+\frac{x^6}6+\cdots[/tex]
(even powers only)
[tex]x^2e^{x^2}=\displaystyle\sum_{n=0}^\infty\frac{x^{2(n+1)}}{n!}=x^2+x^4+\frac{x^6}2+\frac{x^8}6+\cdots[/tex]
(again, even powers only)
So when [tex]n=7[/tex] or [tex]n=9[/tex], the corresponding term in the series is 0, but this is not the case for when [tex]n=8[/tex] or any other even number.
