This seems more like a general request for information than a specific question, so I'll give some analysis, including
- An faster-converging upper bound, which you can use to sandwich the value of $f$,
- The limit of $f(x)$ as $x \to \infty$,
- The power series expansion of $f$.
Also, here is a Desmos graph with a some of the mentioned formulas: https://www.desmos.com/calculator/mlgjz3cu0b
First, note that the term in the sum is positive, so for positive $x$, $f(x)$ is greater than its partial sums. Each partial sum is a lower bound for $f$. We can find an upper bound like this:
\begin{align}
f(x) - \sum_{n = 1}^N \frac{\sin^2(x/n)}x &= \sum_{n = N + 1}^\infty \frac{\sin^2(x/n)}x\\
&\leq \sum_{n = N + 1}^\infty \frac{x^2/n^2}x\\
&= x\sum_{n = N + 1}^\infty \frac1{n^2}\\
&\leq x\int_N^\infty \frac1{t^2}dt\\
&= \frac xN.
\end{align}
Therefore,
$$
\sum_{n = 1}^N \frac{\sin^2(x/n)}x + \frac xN
$$
is an upper bound for $f(x)$. Conveniently, this upper bound converges faster than the original expression, so you can graph it to see a more accurate picture of the function.
Now, I'll show that $f(x) \to \pi/2$ as $x \to \infty$. For every natural number $N$, let
$$
f_N(x) = \sum_{n = 1}^{\lceil Nx\rceil} \frac{\sin^2(x/n)}x.
$$
Clearly, $f_N(x)$ converges pointwise to $f(x)$ as $N \to \infty$. Applying the upper bound we just found, we can see that
$$
|f(x) - f_N(x)| = f(x) - \sum_{n = 1}^{\lceil Nx\rceil} \frac{\sin^2(x/n)}x < \frac x{Nx} = \frac1N,
$$
which implies that the convergence is uniform. This means that we can interchange limits of $x$ and $N$, so
$$
\lim_{x \to \infty} f(x) = \lim_{x \to \infty}\left(\lim_{N \to \infty} f_N(x)\right) = \lim_{N \to \infty}\left(\lim_{x \to \infty} f_N(x)\right).
$$
Now, we will see that $f_N(x)$ is a Riemann sum of $\sin^2(1/t)$, which will allow us to convert it into an integral. Consider the partition $p = \{\frac1x, \frac2x, \frac3x, \dots, \frac{\lceil Nx\rceil + 1}x\}$. Then
$$
f_N(x) = \sum_{n = 1}^{\lceil Nx\rceil} \sin^2\left(\frac1{t^*_n}\right) \Delta t_n,
$$
where $t^*_n = n/x$ and $\Delta t_n = 1/x$. This is a Riemann sum of $\sin^2(1/t)$ over $p$. Since $\sin^2(1/t)$ is Riemann-integrable for positive numbers and $\|p\| = 1/x \to 0$ as $x \to \infty$, we have
$$
\lim_{x \to \infty} f_N(x) = \int_0^N \sin^2\left(\frac1t\right)dt.
$$
This means that
$$
\lim_{N \to \infty}\left(\lim_{x \to \infty} f_N(x)\right) = \int_0^\infty \sin^2\left(\frac1t\right)dt.
$$
Substituting $u = 1/t$ transforms the integral to
$$
\int_0^\infty \frac{\sin^2(u)}{u^2}du,
$$
which is proved to equal $\pi/2$ by the answers to this question. Therefore
$$
\lim_{x \to \infty} f(x) = \frac\pi2.
$$
To find the power series expansion, note that
\begin{align}
f(x) = \sum_{n = 1}^\infty \frac{\sin^2(x/n)}x &= \sum_{n = 1}^\infty \frac{1 - \cos(2x/n)}{2x}\\
&= \sum_{n = 1}^\infty \frac{1 - \left(1 - \frac1{2!}\left(\frac{2x}n\right)^2 + \frac1{4!}\left(\frac{2x}n\right)^4 - \frac1{6!}\left(\frac{2x}n\right)^6 + \cdots\right)}{2x}\\
&= \sum_{n = 1}^\infty \frac{\sum_{k = 1}^{\infty}(-1)^{k-1} \frac1{(2k)!}\left( \frac{2x}n \right)^{2k}}{2x}\\
&=\sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{k-1} \frac1{n^{2k}} \cdot \frac{(2x)^{2k-1}}{(2k)!}.
\end{align}
Because Taylor series converges absolutely, we know that the inner sum converges absolutely. And since the inner sum is $O(1/n^2)$, the double sum must converge absolutely. Therefore, we can swap the sums, or equivalently, swap $n$ and $k$.
\begin{align}
&=\sum_{n=1}^\infty \sum_{k=1}^\infty (-1)^{n-1} \frac1{k^{2n}} \cdot \frac{(2x)^{2n-1}}{(2n)!}\\
&=\sum_{n=1}^\infty (-1)^{n-1} \frac{(2x)^{2n-1}}{(2n)!} \sum_{k=1}^\infty \frac1{k^{2n}}\\
f(x) &= \sum_{n=1}^\infty (-1)^{n-1} \frac{\zeta(2n)(2x)^{2n-1}}{(2n)!}.
\end{align}
