Integration in a non-standard way: evaluating $ \int \frac 1 { x ^ 2 } \, \mathrm d x $ without applying the power rule - Is it nonsense? I want to evaluate the integral
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x $$
in a non-standard way and without applying the power rule.
Let $ x \ne 0 $ and
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = f ( x ) + C \text . $$
For any $ a \in \mathbb R \setminus \{ 0 \} $ we have
$$ f ( a x ) = \int \frac 1 { a ^ 2 x ^ 2 } \, \mathrm d ( a x ) = \frac 1 a \int \frac 1 { x ^ 2 } \, \mathrm d x = \frac 1 a \bigl ( f ( x ) + C \bigr ) \text . $$
Then, we have
$$ f ( x ) = a f ( a x ) \text , $$
for any $ a \in \mathbb R \setminus \{ 0 \} $.
Putting $ x = 1 $, we have
$$ f ( a ) = \frac { f ( 1 ) } a \text . $$
This is the critical point for me. I'm not sure I'm not wrong here. I will just continue.
Here, I will replace the constant $ a $ with the variable $ x $. My argument for doing this is as follows: $ a $ covers all nonzero real numbers.
So, I will write the last equation like this (but, I'm not sure):
$$ f ( x ) = \frac { f ( 1 ) } x \text . $$
We continue.
Let $ u = x $ and $ v = f ( x ) $, so that $ \mathrm d u = \mathrm d x $ and $ \mathrm d v = \frac 1 { x ^ 2 } \, \mathrm d x $.
$$ \int x \cdot \frac 1 { x ^ 2 } \, \mathrm d x = x \cdot f ( x ) - \int f ( x ) \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies x \cdot \frac { f ( 1 ) } x - \int \frac { f ( 1 ) } x \, \mathrm d x = \ln x + C _ 1 \text ; $$
$$ \implies f ( 1 ) - f ( 1 ) \left ( \ln x + C _ 2 \right ) = \ln x + C _ 1 \text ; $$
$$ \implies \bigl ( f ( 1 ) + 1 \bigr ) \ln x = f ( 1 ) ( 1 - C _ 2 ) - C _ 1 \text . $$
The right side is a constant. The left side must also be a constant.
$$ \implies f ( 1 ) + 1 = 0 \implies f ( 1 ) = - 1 \text . $$
Finally, we get
$$ f ( x ) = - \frac 1 x \text ; $$
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = - \frac 1 x + C \text . $$
Remark 1.
My goal was to find antiderivative of $ \frac 1 { x ^ 2 } $. Because,
$$ \int \frac 1 { x ^ 2 } \, \mathrm d x = \text {antiderivative of } \frac 1 { x ^ 2 } + \text{constant.} $$
In short, the function $ f ( x ) $ was an antiderivative.
Remark 2.
Initially $ a $ is a constant. However, it covers all real numbers except zero. Therefore, I replaced it with the variable $ x $.
How much of my work does math allow? Is the method I am using a nonsense? Otherwise, can the errors it contains be corrected?

It's best to use definite integrals; this not only distinguishes certain variables your reasoning risks conflating (though you've done your best to explain why the steps are legal), but also makes it shorter. Define $f(x):=\int_\infty^x\frac{dt}{t^2}$ so $f(ax)=\int_\infty^{ax}\frac{dt}{t^2}=\int_\infty^x\frac{du}{au^2}=\frac{f(x)}{a}$; in particular, $f(a)=\frac{f(1)}{a}$. For $x>0$, integrate by parts with $u:=t,\,v:=f(t)$ so$$\begin{align}\ln x&=\int_1^x\tfrac{dt}{t}\\&=[tf(t)]_1^x-\int_1^xf(t)dt\\&=\underbrace{xf(x)-f(1)}_0-\int_1^x\tfrac{f(1)}{t}dt\\&=-f(1)\ln x\\\implies f(1)&=-1.\end{align}$$
