Differentiation of Taylor Series Let $g(x) = e^{-1/x^2}$ for $x$ not equal to zero, and $g(0) = 0$. 

a) Please Show that $g^{(n)}(0) = 0$, for all $n = 0,1,2,3,4, \ldots$

Can someone please elaborate on the comments below for this one? 

b) Please Show that the Taylor Series for $g$ about 0 agrees with $g$ only at $x = 0$.

I think this would be easy once I have part a, all I have to do is plug in n = 0? 

Can someone please show how to do this?
A good way to do this is to prove something stronger.  Let $P$ be any polynomial.  Then show that the function $h$ defined by
$h(x) = P(1/x)e^{-1/x^2}$ for $x \ne 0$ and $h(x)=0$ has your property $h^{(n)}(0) = 0$ for all $n$.  Do it by showing: (1) $h$ is continuous; and (2) the derivative of such a function $h$ is another function of the same kind.
All you have to do is prove that $\dfrac{\rm{d}g(x)}{\rm{d}x} = (\text{something})\,g(x)$ so that the limit of any derivative when $x\rightarrow 0$ is zero, since g(0)=0.

I see that

$$\dfrac{\rm{d}g(x)}{\rm{d}x} = \dfrac{2}{x^3}\,g(x) $$