= Solution
Use the usual probability-kernel convention: $K\ge0$, $\int K=1$, and $K\in L^2$, in addition to symmetry and <compact support>. These conditions are needed for the stated finite bound; symmetry and <compact support> by themselves do not impose normalization or nonnegativity. Interpret the differentiability assumption in the usual <Sobolev space> sense $f\in H^2(\mathbb R)$, or assume classical regularity sufficient for the integral <Taylor remainder>. The <kernel density estimator> is
$$
\widehat f_h(x)=\frac1{nh}\sum_{i=1}^nK\!\left(\frac{x-X_i}{h}\right),\qquad K_h(x)=h^{-1}K(x/h),
\qquad \mathbb E\widehat f_h=K_h*f.
$$
The <bias-variance decomposition of mean squared error>, followed by <Tonelli theorem>, yields
$$
\mathbb E\|\widehat f_h-f\|_2^2
=\int\operatorname{Var}(\widehat f_h(x))\,dx+\|K_h*f-f\|_2^2.
$$
<Independence> and a <change of variables> give the exact <integrated variance of a kernel density estimator>
$$
\int\operatorname{Var}(\widehat f_h(x))\,dx
=\frac1n\left(\frac{\|K\|_2^2}{h}-\|K_h*f\|_2^2\right)
\le\frac{\|K\|_2^2}{nh}.
$$
All changes of integration order are justified either by nonnegativity or by the displayed finite integrals.
For the <bias of a kernel density estimator>, symmetry gives $\int uK(u)\,du=0$. Taylor's integral remainder, with translations understood in $L^2$, gives
$$
f(x-hu)-f(x)+hu f'(x)=h^2u^2\int_0^1(1-t)f''(x-thu)\,dt.
$$
The <Minkowski inequality> and translation invariance of the <L2 norm> imply
$$
\|f(\cdot-hu)-f+hu f'\|_2\le\frac{h^2u^2}{2}\|f''\|_2,
\qquad
\|K_h*f-f\|_2\le\frac{h^2}{2}\mu_2(K)\|f''\|_2,
$$
where $\mu_2(K)=\int u^2K(u)\,du$. Consequently we obtain the slightly stronger conclusion
$$
\boxed{\mathbb E\|\widehat f_h-f\|_2^2\le
\frac{\|K\|_2^2}{nh}+\frac14h^4\mu_2(K)^2\|f''\|_2^2.}
$$
Since $1/4\le1/3$, this proves the requested bound. For signed kernels, the same argument has $\int u^2|K(u)|\,du$ in place of $\mu_2(K)$; the printed bound is not generally justified merely by signed-kernel symmetry. For example, let $U_a=(2a)^{-1}\mathbf1_{[-a,a]}$ and $K=(4/3)U_1-(1/3)U_2$. This symmetric compactly supported signed kernel has integral one and $\mu_2(K)=0$, but its fourth moment is nonzero. For a normal density its convolution has nonzero bias at any fixed positive bandwidth. Sending $n\to\infty$ would therefore contradict a bound containing only the integrated variance term.
Writing the requested bound as $A/(nh)+Bh^4$, differentiation gives $h_*=(A/(4Bn))^{1/5}$ when $B>0$. Thus \b[$h\asymp n^{-1/5}$ gives <mean integrated squared error> of order $n^{-4/5}$]. With the sharper constant one instead has $h_*=(\|K\|_2^2/(n\mu_2(K)^2\|f''\|_2^2))^{1/5}$; the rate is identical.
A direct heuristic for $J_2(f)=\|f''\|_2^2$ is to choose a twice differentiable pilot kernel $L$, estimate $f''$ by the <second derivative kernel density estimator>
$$
\widehat f_b''(x)=\frac1{nb^3}\sum_i L''\!\left(\frac{x-X_i}{b}\right),
$$
and integrate its square. Its integrated <variance> contains a diagonal term $\|L''\|_2^2/(nb^5)$, which should not be mistaken for the target. A useful corrected <U-statistic> removes that diagonal:
$$
\widehat J_{2,b}=\frac1{n(n-1)b^5}\sum_{i\ne j}A\!\left(\frac{X_i-X_j}{b}\right),
\qquad A(v)=\int L''(t)L''(t+v)\,dt.
$$
Its <expectation> is $\|L_b*f''\|_2^2$. Thus smoothing <bias of an estimator>, rather than the diagonal contribution, remains. For a smooth compactly supported probability pilot kernel, the same variance calculation gives
$$
\mathbb E\|\widehat f_b''-f''\|_2^2\le\frac{\|L''\|_2^2}{nb^5}+\|L_b*f''-f''\|_2^2.
$$
The second term tends to zero because convolution with a shrinking probability kernel approximates every $L^2$ function. Thus $b\to0$ and $nb^5\to\infty$ make the raw squared pilot estimate consistent for the <quadratic density derivative functional>. If $Q_b=\int(\widehat f_b'')^2$, then
$$
\widehat J_{2,b}=\frac n{n-1}\left(Q_b-\frac{\|L''\|_2^2}{nb^5}\right),
$$
so the diagonal-corrected <U-statistic> is consistent under the same choice. Consistency already uses the assumed $f''\in L^2$; the additional third-derivative hypothesis is relevant to more ambitious rate heuristics.
\b[Three bounded continuous <derivatives> do not justify a general root-$n$ estimation rate for $J_2$.] For a smooth perturbation $a$ with <compact support>, the <directional derivative> is
$$
\left.\frac{d}{dt}J_2(f+ta)\right|_{t=0}
=2\int f''a''=2\int f^{(4)}a,
$$
where the last expression requires a fourth <derivative> in the appropriate weak sense. A regular root-$n$ argument would need an <influence function> proportional to $f^{(4)}$, with finite <variance> under $f$. Three <derivatives> do not supply that condition.
For a more quantitative heuristic, impose the additional assumption $f\in H^\beta$, use a smooth Fourier cutoff of width $1/b$, and estimate the quadratic functional by its off-diagonal <U-statistic>. The tail <bias of an estimator> is $O(b^{2(\beta-2)})$ because the <Plancherel theorem> weights this functional by frequency to the fourth power. The degenerate part of its standard deviation is $O(n^{-1}b^{-9/2})$: the squared $L^2$ norm of the quadratic kernel scales as $b^{-9}$. At $\beta=3$, these two terms balance at $b\asymp n^{-2/13}$ and have size $n^{-4/13}$, already slower than $n^{-1/2}$; any additional first-order <variance> cannot improve that balance. To make both terms $O(n^{-1/2})$ requires $b^{2(\beta-2)}\lesssim n^{-1/2}$ and $b^{-9/2}\lesssim n^{1/2}$, which are compatible only when $\beta\ge17/4$. This is an illustrative smoothness calculation under extra Sobolev and moment assumptions, not a rate theorem asserted solely from bounded third <derivatives>. Particular much smoother <probability density functions> may of course permit root-$n$ estimation.
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