= Solution
Put
$$
W(n)=\sum_{d\mid n}\mu(d)f\left(\frac{\log d}{\log D}\right).
$$
If $Y<n\leq Y+X$ is prime, then $n>D=X^{1/10}$. Since $f$ is supported on $[-1,1]$, the only divisor of $n$ contributing to $W(n)$ is $d=1$, and $W(n)=f(0)=1$. Every summand $W(n)^2$ is nonnegative, so
$$
\sum_{Y<n\leq Y+X}W(n)^2\geq\pi(Y+X)-\pi(Y).
$$
Let
$$
z_j=\frac{1-2\pi it_j}{\log D}.
$$
Since $g$ is the <Fourier transform> of $x\mapsto e^xf(x)$, the <Fourier inversion theorem> gives the <Fourier representation of a smooth Selberg weight>
$$
f\left(\frac{\log d}{\log D}\right)
=\int_{-\infty}^{\infty}g(t)d^{-(1-2\pi it)/\log D}\,dt.
$$
Expanding the square, interchanging the absolutely convergent sums and integrals, and using
$$
\#\{Y<n\leq Y+X:[d_1,d_2]\mid n\}
=\frac X{[d_1,d_2]}+O(1),
$$
we obtain
$$
\sum_{Y<n\leq Y+X}W(n)^2
=X\iint_{\mathbb R^2}g(t_1)g(t_2)H(t_1,t_2,D)\,dt_1dt_2+O(D^2).
$$
The error is $O(D^2)$ because the support of $f$ restricts both divisors to $d_j\leq D$. The main factor is
$$
H(t_1,t_2,D)
=\sum_{d_1,d_2\geq1}
\frac{\mu(d_1)\mu(d_2)}{[d_1,d_2]d_1^{z_1}d_2^{z_2}},
$$
and its <Euler product> is
$$
\boxed{
H(t_1,t_2,D)
=\prod_p\left(
1-p^{-1-z_1}-p^{-1-z_2}+p^{-1-z_1-z_2}
\right).}
$$
It remains to estimate the integral using the assumed zeta-factor bound. The transform $g$ is a <Schwartz function>, since $e^xf(x)$ is smooth and compactly supported. We may therefore truncate to $|t_1|,|t_2|\leq T$, losing an arbitrarily large negative power of $T$. Uniformly in the needed truncated range, the standard estimates near the pole of the <Riemann zeta function> give
$$
\left|\zeta\left(1+\frac{1-2\pi it}{\log D}\right)^{-1}\right|
\ll\frac{1+|t|}{\log D}
$$
and
$$
\left|\zeta\left(1+\frac{2-2\pi i(t_1+t_2)}{\log D}\right)\right|
\ll\frac{\log D}{1+|t_1+t_2|}+O(\log(2+T)).
$$
After multiplication, one net factor $(\log D)^{-1}$ remains. The polynomial factors in $t_1,t_2$ are integrable against the rapidly decreasing $g(t_1)g(t_2)$, and choosing $T$ as a sufficiently large power of $\log D$ makes the tails negligible. Hence
$$
\iint_{\mathbb R^2}|g(t_1)g(t_2)H(t_1,t_2,D)|\,dt_1dt_2
\ll\frac1{\log D}.
$$
Since $D=X^{1/10}$ and $D^2=X^{1/5}\ll X/\log X$,
$$
\boxed{\pi(Y+X)-\pi(Y)\ll\frac X{\log D}\ll\frac X{\log X}.}
$$
This proves the <short-interval prime upper bound from a smooth divisor weight>.
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