Let $g(x)=e^{−1/x^2}$ for $x\neq0$ and $g(0)=0$.
(a) Show $g^{(n)}(0)=0$ for all $n = 0,1,2,3,\dots$
(b) Show the Taylor series for $g$ about $0$ agrees with $g$ only at $x = 0$.
So far I showed that there exists a polynomial $p_n$ of degree $3n$ such that $$g^{(n)}=e^{−1/x^2}p_n\left(\frac{1}{x}\right).$$ Then I found the summation formula for it and took the derivative. And that shows that the polynomial $p_n$ holds for all $n$. I'm not sure about the next few steps I have to take. I am guessing that using induction and the definition of a limit I check if $g^{(n)}(0)=0$ for $n=0$ and then check if it holds for $n+1$? Is this reasoning correct and how would I check it for $n+1$?
Yes, use induction. Given that $g^{(n)}(0)=0$ write $g^{(n+1)}(0)=0$ as the linit of $\frac {g^{n} (x)} x$ as $x \to 0$. So you have to show that $e^{-1/x^{2}} \frac 1 x p(\frac 1 x) \to 0$. Clerly this will follow if you show that $e^{-1/x^{2}} \frac 1 {x^{n}} \to 0$ for each positive integer $n$. Equivalently, you have to show that $e^{-y^{2}}y^{n} \to 0$ as $y \to \infty$. For this write $e^{-y^{2}}y^{n}$ as $\frac {y^{n}} {e^{y^{2}}}$. This has the indeterminat form $\frac {\infty} {\infty}$. Apply L'hopital's Rule $n$ times to complete the proof.