Let $Z \sim \mathcal N(0,1)$ be a standard normally distributed random variable and let $f: \mathbb R \to \mathbb R$ be a differentiable function such that $f'(Z)$ and $f(Z) Z$ are in $\mathcal L^1$.
I need to show that $\mathbb E[f'(Z)] = \mathbb E[Z f(Z)],$ preferably using partial integration.
By applying the LOTUS rule, I have already shown that $$ \mathbb E [Zf(Z)] = \int_\mathbb R f(x) x \rho_Z(x) dx $$ where $\rho_Z(x) = \frac 1 {\sqrt{2\pi}} e^{- \frac x 2}$, i.e., the pdf of $Z$. Using partial integration on the l.h.s. yields $$ \mathbb E[f'(Z)] = f(x) \rho_Z(x) \Big|_{x=-\infty}^{x=\infty} + \int_\mathbb R f(x) x \rho_Z(x) dx $$ thus it remains to show (if I haven't made a mistake) that $f(x) \rho_Z(x) \Big|_{x=-\infty}^{x=\infty} = 0$ which I am skeptical of.
Hints: $$ \mathbb E[f'(Z)I_{|Z|\leq n}] = f(x) \rho_Z(x) \Big|_{x=-n}^{x=n} + \int_{[-n,n]} f(x) x \rho_Z(x) dx .$$ Since the left side as well as the second term on the right have a finite limit as $n \to \infty$ it follows that $f(x) \rho_Z(x) \Big|_{x=-n}^{x=n}$ also has finite limit as $n \to \infty$. If this limit is not $0$ you can easily check that $E|Zf(Z)| =\infty$.