$f(x)=\lim_{n\to\infty}\left(\frac{n^n}{n!}\prod_{r=1}^n\frac{x^2+\frac{n^2}{r^2}}{x^3+\frac{n^3}{r^3}}\right)^{\frac xn}$

Question

$$f(x)=\lim_{n\to\infty}\left(\frac{n^n}{n!}\prod_{r=1}^n\frac{x^2+\frac{n^2}{r^2}}{x^3+\frac{n^3}{r^3}}\right)^{\frac xn}$$What is the monotonicity of $f(x),x\gt0$

I tried in the following way [METHOD-$1$]:

$$f(x)=\lim_{n\to\infty}\left(\frac{n^n}{n!}\prod_{r=1}^n\frac{x^2+\frac{n^2}{r^2}}{x^3+\frac{n^3}{r^3}}\right)^{\frac xn} \implies \lim_{n\to\infty}\left(\frac{1}{n!}\prod_{r=1}^n\frac{\left(\frac xn\right)^2+\frac{1}{r^2}}{\left(\frac xn\right)^3+\frac{1}{r^3}}\right)^{\frac xn}\qquad......... (1)$$ Now for any finite $x$, $$\lim_{n\to\infty} \frac xn \to 0$$ This gives, $f(x)= \left( \left(\frac 1{n!}\right)^2 \over n!.\left(\frac 1{n!}\right)^3\right)^0 \implies (1)^0 = 1$ $\implies f(x)=1$

But here is another approach [METHOD-$2$]:

Continuing from $(1)$, Taking natural log both sides gives; [let $f(x)=y$] $$ln(y) = \lim_{n\to\infty}\frac xn \sum_{r=1}^n ln\left(\frac{\left(\frac xn\right)^2+\frac{1}{r^2}}{\left(\left(\frac xn\right)^3+\frac{1}{r^3}\right)r}\right) \implies \lim_{n\to\infty}ln\left(\frac xn \sum_{r=1}^n \frac{\left(\frac {xr}n\right)^2+1}{\left(\frac {xr}n\right)^3+1}\right)$$ Taking, $\frac rn = t$ and $\frac 1n=dt$; $$\implies ln(y)=\int_0^1ln\left(1+(xt)^2\over1+(xt)^3\right).(xdt)$$ Now $xt=k$ and $xdt=dk$, $$ln(y)=\int_0^xln\left(1+(k)^2\over1+(k)^3\right).(dk)$$ Differentiating both sides, $$\frac{y'}y=ln\left(1+(k)^2\over1+(k)^3\right)$$

Now since $y\ge0$;

$y'\ge0$ for $0\lt x\le1$

$y'\le0$ for $x\ge1$

So $f(x)$ is increasing for $0\lt x\lt1$ and $f(x)$ is decreasing for $x\gt1$

The answer in the 2 cases do not match.

And the answer in METHOD-$2$ happens to be the correct answer.

I need help to find what is wrong with METHOD-$1$