= Solution
Write $R(v)=\int_{\mathbb R}v(u)^2\,du$. The <kernel density estimator> with bandwidth $h>0$ is
$$
\widehat f_h(x)=\frac1{nh}\sum_{i=1}^nK\!\left(\frac{x-X_i}{h}\right).
$$
Sufficient conditions are that $K$ be a nonnegative symmetric <probability density function> with $R(K)<\infty$ and finite nonzero <second moment> $\mu_2(K)=\int u^2K(u)\,du$, and that $h\to0$, $nh\to\infty$. Symmetry and the second-moment assumption give $\int uK(u)\,du=0$. The <expectation> is a <convolution>:
$$
\mathbb E\widehat f_h(x)=\int K(u)f(x-hu)\,du.
$$
<Taylor's theorem> with integral remainder gives
$$
f(x-hu)=f(x)-hu f'(x)+h^2u^2\int_0^1(1-s)f''(x-shu)\,ds.
$$
The zeroth-order term integrates to $f(x)$, and the <first-order term> vanishes. The <second derivative> is bounded and continuous, so <dominated convergence>, with dominating function proportional to $u^2K(u)$, yields
$$
\boxed{\operatorname{Bias}\{\widehat f_h(x)\}=\tfrac12h^2\mu_2(K)f''(x)+o(h^2).}
$$
Combining this with the supplied leading <variance> gives the <asymptotic mean squared error>
$$
\operatorname{AMSE}_x(h)=\frac{R(K)f(x)}{nh}+\frac{\mu_2(K)^2f''(x)^2h^4}4.
$$
When $f(x)>0$ and $f''(x)\ne0$, its derivative is $-R(K)f(x)/(nh^2)+\mu_2(K)^2f''(x)^2h^3$. It changes sign exactly once, giving the <pointwise optimal kernel bandwidth>
$$
\boxed{h_{\rm AMSE}(x)=\left\{\frac{R(K)f(x)}{n\mu_2(K)^2f''(x)^2}\right\}^{1/5}.}
$$
\b[Positivity of $f(x)$ is needed for this positive interior optimum.] The printed condition $f''(x)\ne0$ alone does not ensure it. For instance, $f(x)=x^2\phi(x)$ is a normalized bounded <probability density function> with bounded continuous square-integrable <second derivative>, but $f(0)=0$ and $f''(0)=2\phi(0)>0$. At this point the displayed leading <variance> vanishes and the two-term AMSE has no positive interior minimizer; finer <variance> terms are needed to determine an appropriate bandwidth. The formal formula there gives zero, which does not satisfy the bandwidth assumptions.
For integrated error, the same expansion holds in $L^2$: translation <continuity> of $f''\in L^2$ applied to the integral remainder gives
$$
\|\mathbb E\widehat f_h-f-\tfrac12h^2\mu_2(K)f''\|_2=o(h^2).
$$
Indeed Minkowski's integral inequality bounds the norm of the difference, divided by $h^2$, by the integral of $u^2K(u)\int_0^1(1-s)\|f''(\cdot-shu)-f''\|_2\,ds\,du$; each translated difference tends to zero and is bounded by $2\|f''\|_2$. This proves the claimed integrated <bias expansion> without assuming a uniform pointwise remainder over the whole <real line>.
If $K_h(x)=h^{-1}K(x/h)$, direct integration of the <variance> gives
$$
\int\operatorname{Var}\{\widehat f_h(x)\}\,dx=\frac{R(K)}{nh}-\frac{R(K_h*f)}n.
$$
The last term is $O(n^{-1})$ because $f$ is bounded and integrable, hence square integrable, and <convolution> with a <probability density function> does not increase its $L^2$ norm. It is negligible compared with $1/(nh)$. Thus the <asymptotic mean integrated squared error> and its minimizer are
$$
\operatorname{AMISE}(h)=\frac{R(K)}{nh}+\frac{\mu_2(K)^2R(f'')h^4}4,\qquad \boxed{h_{\rm AMISE}=\left\{\frac{R(K)}{n\mu_2(K)^2R(f'')}\right\}^{1/5}.}
$$
Here $R(f'')>0$: otherwise the continuous <second derivative> vanishes everywhere, making $f$ affine, impossible for a <probability density function> on the whole <real line>.
Now take the <standard normal density> $f=\phi$. Differentiation gives $\phi''(x)=(x^2-1)\phi(x)$. Gaussian integration gives
$$
R(\phi'')=\frac1{2\pi}\int(x^2-1)^2e^{-x^2}\,dx=\frac3{8\sqrt\pi}.
$$
The kernel constants cancel from the bandwidth ratio. At $x\ne\pm1$,
$$
\left(\frac{h_{\rm AMSE}(x)}{h_{\rm AMISE}}\right)^5=\frac{\phi(x)R(\phi'')}{\phi''(x)^2}=\frac{3\sqrt2}{8}\frac{e^{x^2/2}}{(x^2-1)^2}.
$$
Put $t=x^2$. The <logarithmic derivative> of $e^{t/2}/(t-1)^2$ is $1/2-2/(t-1)$. On $0\leq t<1$ the function increases, giving its minimum at $t=0$, with value one. On $t>1$ it decreases up to $t=5$ and then increases, giving its minimum $e^{5/2}/16<1$. Thus the <global minimizers> are $x=\pm\sqrt5$, and the <normal-density local-to-global bandwidth ratio> is
$$
\boxed{\inf_{x\ne\pm1}\frac{h_{\rm AMSE}(x)}{h_{\rm AMISE}}=\left(\frac{3\sqrt2 e^{5/2}}{128}\right)^{1/5}=\left(\frac{9e^5}{8192}\right)^{1/10}.}
$$
The ratio diverges at the two <inflection points> $\pm1$, where the second-order pointwise bias approximation vanishes and needs a higher-order replacement.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-42-bandwidth-ratio.png]
{title=Normal-density pointwise bandwidth relative to the integrated-error optimum; minima at plus and minus square root of five}
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