1. Rewrite the product neatly

Given

$$f(x)=\lim_{n\to\infty}\left(\frac{n^n\,(x+n)(x+n/2)\cdots (x+n/n)}{n!\,(x^2+n^2)(x^2+n^2/4)\cdots (x^2+n^2/n^2)}\right)^{x/n},\qquad x>0.$$

Observe that

$$(x+n/k)=\frac{n}{k}\left(1+\frac{kx}{n}\right),$$

and

$$\left(x^2+\frac{n^2}{k^2}\right)=\frac{n^2}{k^2}\left(1+\frac{k^2x^2}{n^2}\right).$$

Hence

$$\prod_{k=1}^n (x+n/k)=\prod_{k=1}^n \frac{n}{k}\prod_{k=1}^n\left(1+\frac{kx}{n}\right) =\frac{n^n}{n!}\prod_{k=1}^n\left(1+\frac{kx}{n}\right),$$

and

$$\prod_{k=1}^n \left(x^2+\frac{n^2}{k^2}\right) =\prod_{k=1}^n \frac{n^2}{k^2}\prod_{k=1}^n\left(1+\frac{k^2x^2}{n^2}\right) =\frac{n^{2n}}{(n!)^2}\prod_{k=1}^n\left(1+\frac{k^2x^2}{n^2}\right).$$

Substitute into the expression inside the limit:

$$\frac{n^n\prod_{k=1}^n (x+n/k)}{n!\prod_{k=1}^n (x^2+n^2/k^2)} = \frac{n^n\cdot \frac{n^n}{n!}\prod_{k=1}^n(1+kx/n)}{n!\cdot \frac{n^{2n}}{(n!)^2}\prod_{k=1}^n(1+k^2x^2/n^2)} =\prod_{k=1}^n \frac{1+\frac{kx}{n}}{1+\frac{k^2x^2}{n^2}}.$$

Now use

$$1+\frac{k^2x^2}{n^2}=\left(1+\frac{kx}{n}\right)\left(1-\frac{kx}{n}+\frac{k^2x^2}{n^2}\right).$$

So each factor becomes

$$\frac{1+\frac{kx}{n}}{1+\frac{k^2x^2}{n^2}}=\frac{1}{1-\frac{kx}{n}+\frac{k^2x^2}{n^2}}.$$

Therefore

$$f(x)=\lim_{n\to\infty}\left(\prod_{k=1}^n \frac{1}{1-\frac{kx}{n}+\frac{k^2x^2}{n^2}}\right)^{x/n}.$$

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2. Take logarithm and convert to an integral

Let

$$\ln f(x)=\lim_{n\to\infty} \frac{x}{n}\sum_{k=1}^n \ln\left(\frac{1}{1-\frac{kx}{n}+\frac{k^2x^2}{n^2}}\right).$$

Thus

$$\ln f(x)= -x\lim_{n\to\infty}\frac1n\sum_{k=1}^n \ln\left(1-x\frac{k}{n}+x^2\frac{k^2}{n^2}\right).$$

This is a Riemann sum, so

$$\ln f(x)= -x\int_0^1 \ln(1-xt+x^2t^2)\,dt.$$

Now set $u=xt$, so $dt=du/x$. Then

$$\ln f(x)= -\int_0^x \ln(1-u+u^2)\,du.$$

Hence

$$f(x)=\exp\left(-\int_0^x \ln(1-u+u^2)\,du\right).$$

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3. Differentiate

Let

$$F(x)=\ln f(x)=-\int_0^x \ln(1-u+u^2)\,du.$$

Then by FTC,

$$F'(x)=\frac{f'(x)}{f(x)}=-\ln(1-x+x^2).$$

So

$$f'(x)= -f(x)\ln(1-x+x^2).$$

Since $f(x)>0$ for all $x>0$, the sign of $f'(x)$ is determined by $-\ln(1-x+x^2)$.

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4. Check option C

At $x=2$,

$$1-x+x^2=1-2+4=3.$$

Therefore

$$\frac{f'(2)}{f(2)}=-\ln 3<0$$

and since $f(2)>0$,

$$f'(2)<0.$$

So

$$f'(2)\le 0$$

is true.

Thus C is correct.

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5. Check option D

We have

$$\frac{f'(x)}{f(x)}=-\ln(1-x+x^2).$$

So compare at $x=3$ and $x=2$:

$$\frac{f'(3)}{f(3)}=-\ln(7),\qquad \frac{f'(2)}{f(2)}=-\ln(3).$$

Since $\ln 7>\ln 3$,

$$-\ln 7< -\ln 3.$$

Hence

$$\frac{f'(3)}{f(3)}<\frac{f'(2)}{f(2)}.$$

So option D is false.

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6. Study monotonicity for options A and B

Since

$$\frac{f'(x)}{f(x)}=-\ln(1-x+x^2),$$

we inspect $1-x+x^2$.

Note

$$1-x+x^2=\left(x-\frac12\right)^2+\frac34.$$

Also,

$$1-x+x^2\le 1$$

when

$$x^2-x\le 0 \iff x\in[0,1].$$

Thus for $0<x<1$,

$$\ln(1-x+x^2)\le 0 \quad\Rightarrow\quad -\ln(1-x+x^2)\ge 0.$$

Therefore $f'(x)\ge 0$ on $(0,1)$, with strict positivity except at $x=1$? Check:

$$1-x+x^2=1 \iff x(x-1)=0.$$

So on $(0,1)$ except endpoints, actually $1-x+x^2<1$, hence $f'(x)>0$. Thus $f$ is increasing on $(0,1)$.

Therefore:

• Since $\frac12<1$, $f(1/2)<f(1),$ so option A is false.
• Since $\frac13<\frac23$, $f(1/3)<f(2/3),$ hence $f(1/3)\le f(2/3),$ so option B is true.

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7. Final selection

Correct options are

$$\boxed{B,\ C}.$$

This matches the stored correct answer.