Elementary calculus estimate or not?

This really belongs to MSE rather than to MO, but I'm too lazy to initiate the moving process, so I'll just answer.

There may be more intelligent ways to do it, but you can also integrate by parts and get what you want, say, for smooth functions with compact support, after which you should carefully pass to the limit to extend it to the corresponding Sobolev class.

Let's just solve a more general problem. For non-negative integer $a,b,c$ and an infinitely smooth compactly supported real-valued $f$, denote
$$ I(a,b,c)=\int_{-\infty}^\infty x^af^{(b)}(x)f^{(c)}(x)dx. $$

Claim: If $r,R>0$ are integers and $b,c\le R$, $\frac ar+\frac{b+c-2R}{R}\le 0$, then $$|I(a,b,c)|\le C[I(0,0,0)+I(2r,0,0)+I(0,R,R)]$$

Indeed, there are only finitely many triples $(a,b,c)$ with the above property (admissible triples). Let $M$ be the maximum of $|I(a,b,c)|$ over the admissible triples. The integration by parts formula yields $$ |I(a,b,c)|\le a|I(a-1,b-1,c)|+|I(a,b-1,c+1)| $$ as long as $b>0$ and $c<R$. Note that the triples on the right are still admissible (if $a=0$, the first term just disappears) and $b$ gets smaller every time we do this trick. Thus, after finitely many operations, we arrive at a bound including only "extremal admissible triples" where either $b=0$ or $c=R$. Let $m$ be the maximum of $|I(a,b,c)|$ over the extremal admissible triples. We see that $M\le C_1m$ for some $C_1>0$.

If the extremal admissible triple is of the kind $(a,b,R)$, then we can use Cauchy-Schwarz to write $$ |I(a,b,R)|\le \delta I(2a,b,b)+\delta^{-1}I(0,R,R) $$ with any $\delta>0$ we want. Note that the triple $(2a,b,b)$ is still admissible if $(a,b,R)$ is. We'll choose $\delta=(2C_1)^{-1}$. Then, if $m$ is attained on a triple of this type, we have $$ M\le C_1m\le C_1[\delta M+\delta^{-1}I(0,R,R)]=\frac M2+2C_1^2I(0,R,R) $$ whence $M\le 4C_1^2I(0,R,R)$.

Let us now consider extremal admissible triples of the kind $(a,0,c)$. If $a\le r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(0,c,c)+\delta^{-1}I(2a,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2a,0,0)\le 4C_1^2[I(0,0,0)+I(2r,0,0)] $$ (the crucial point is that $(0,c,c)$ is admissible).

If $a>r$, then we can use Cauchy-Schwarz in the form $$ |I(a,0,c)|\le \delta I(2(a-r),c,c)+\delta^{-1}I(2r,0,0) $$ and, with the same choice of $\delta$ and same argument, get $$ M\le 4C_1^2I(2r,0,0) $$ (the crucial point is that $(2(a-r),c,c)$ is now admissible).

Either way, we get what we claimed. The inequality the OP asked about corresponds to the case $r=R=3$, $a=b=c=2$.

The inequality is true. One integration by parts and standard Cauchy-Schwarz gives $$\int |xf''|^2 \leq C\left(\int |f'''|^2 + x^4|f'|^2 + |f'|^2\right).$$ The second term can be handled by integrating by parts once and applying Cauchy-Schwarz: $$\int x^4|f'|^2 \leq \delta \int x^2|f''|^2 + C_{\delta}\left(\int(x^2+x^6)f^2\right),$$ and $x^2+x^6$ is smaller than a fixed constant times $(1+x^6)$. Finally, the third term in the first inequality can be handled using the Fourier transform and Parseval's identity: $$\int |f'|^2 = \int |\xi|^2|\hat{f}|^2 \leq \int(1+|\xi|^6)|\hat{f}|^2 = \int (|f|^2 + |f'''|^2).$$