Is $x=0$ the inflection point of $f$? 
Is $x=0$ the inflection point of $f= \begin{cases} \sin \frac{1}{x}\cdot e^{\frac{-1}{x^2}}, x \neq 0 \\ 0, x=0 \end{cases}?$

I know that the infection point exist when $f''$ changes the sign in this point.I examine that $f$ has $f', f''$ in $x=0$ but from Mathematica I know that $$f''=\begin{cases} -\frac{8 e^{-\frac{1}{x^2}} \left(\frac{\sin \left(\frac{1}{x}\right)}{x^7}-\frac{5 \cos \left(\frac{1}{x}\right)}{x^6}\right)}{x^3}-\frac{2 \left(\frac{4 e^{-\frac{1}{x^2}}}{x^6}-\frac{6 e^{-\frac{1}{x^2}}}{x^4}\right) \cos \left(\frac{1}{x}\right)}{x^5}-2 e^{-\frac{1}{x^2}} \left(-\frac{10 \sin \left(\frac{1}{x}\right)}{x^8}+\frac{30 \cos \left(\frac{1}{x}\right)}{x^7}+\frac{-\frac{\cos \left(\frac{1}{x}\right)}{x^4}-\frac{2 \sin \left(\frac{1}{x}\right)}{x^3}}{x^5}\right), x \neq 0 \\ 0, x=0 \end{cases}$$ so it is really labor-intensive to calculate  $f''$ and then calculate when $f''>/<0$.Do you know any faster way to do this?

$f$ is of the form
$$\tag1f(x)=\begin{cases}\left(p(\frac1x)\sin\frac1x+q(\frac1x)\cos\frac1x\right)e^{-1/{x^2}},&x\ne0\\0,&x=0 \end{cases} $$
where $p,q$ are polynomials. Show that every function of form $(1)$ is differentiable with derivative also of form $(1)$ (but of course with diffferent polynomials $p,q$). 
Also, unless both $p$ and $q$ are the zero polynomial, for every function of form $(1)$, there exist sequences  $x_n\to 0^+$ and $y_n\to 0^-$such that $f$ changes sign at the $x_n$ and at the $y_n$. Conclude that such $f$ assumes positive and negative values near $x=0$ in such a way that $f$ neither changes sign at $x=0$ nor has an extremum at $x=0$.
Now with the specific $f$ of your porblem statement, apply this result to $f''$ and conclude that $f$ does not have an inflection point at $x=0$.
